Power Series

Section 3.5 Power Series

Let's return to the simple geometric series

\begin{gather*} \sum_{n=0}^\infty x^n \end{gather*}

where $x$ is some real number. As we have seen (back in Example 3.2.4 and Lemma 3.2.5), for $|x| \lt 1$ this series converges to a limit, that varies with $x\text{,}$ while for $|x|\geq 1$ the series diverges. Consequently we can consider this series to be a function of $x$

\begin{align*} f(x) &= \sum_{n=0}^\infty x^n & \text{on the domain $|x| \lt 1$}. \end{align*}

Furthermore (also from Example 3.2.4 and Lemma 3.2.5) we know what the function is.

\begin{align*} f(x) &= \sum_{n=0}^\infty x^n = \frac{1}{1-x}. \end{align*}

Hence we can consider the series $\sum_{n=0}^\infty x^n$ as a new way of representing the function $\frac{1}{1-x}$ when $|x| \lt 1\text{.}$ This series is an example of a power series.

Of course, representing a function as simple as $\frac{1}{1-x}$ by a series doesn't seem like it is going to make life easier. However the idea of representing a function by a series turns out to be extremely helpful. Power series turn out to be very robust mathematical objects and interact very nicely with not only standard arithmetic operations, but also with differentiation and integration (see Theorem 3.5.13). This means, for example, that

\begin{align*} \diff{}{x} \left\{\frac{1}{1-x}\right\} &= \diff{}{x} \sum_{n=0}^\infty x^n & \text{provided $|x| \lt 1$}\\ &= \sum_{n=0}^\infty \diff{}{x} x^n & \text{just differentiate term by term}\\ &= \sum_{n=0}^\infty n x^{n-1}\\ \end{align*}

and in a very similar way

\begin{align*} \int \frac{1}{1-x} \dee{x} &= \int \sum_{n=0}^\infty x^n \dee{x} & \text{provided $|x| \lt 1$}\\ &= \sum_{n=0}^\infty \int x^n \dee{x} & \text{just integrate term by term}\\ &= C + \sum_{n=0}^\infty \frac{1}{n+1} x^{n+1} \end{align*}

We are hiding some mathematics under the word “just” in the above, but you can see that once we have a power series representation of a function, differentiation and integration become very straightforward.

So we should set as our goal for this section, the development of machinery to define and understand power series. This will allow us to answer questions ¹ like

\begin{align*} \text{Is }\ e^x &=\sum\limits_{n=0}^\infty\frac{x^n}{n!} \text{ ? } \end{align*}

Our starting point (now that we have equipped ourselves with basic ideas about series), is the definition of power series.

Subsection 3.5.1 Radius and Interval of Convergence

Definition 3.5.1.

A series of the form

\begin{equation*} A_0 +A_1(x-c) + A_2(x-c)^2 + A_3 (x-c)^3 + \cdots =\sum_{n=0}^\infty A_n(x-c)^n \end{equation*}

is called a power series in $(x-c)$ or a power series centered on $c$. The numbers $A_n$ are called the coefficients of the power series.

One often considers power series centered on $c=0$ and then the series reduces to

\begin{equation*} A_0 +A_1 x + A_2 x^2 + A_3 x^3 + \cdots =\sum_{n=0}^\infty A_n x^n \end{equation*}

For example $\sum_{n=0}^\infty \frac{x^n}{n!}$ is the power series with $c=0$ and $A_n=\frac{1}{n!}\text{.}$ Typically, as in that case, the coefficients $A_n$ are given fixed numbers, but the “$x$” is to be thought of as a variable. Thus each power series is really a whole family of series — a different series for each value of $x\text{.}$

One possible value of $x$ is $x=c$ and then the series reduces ² to

\begin{align*} \sum_{n=0}^\infty A_n (x-c)^n\Big|_{x=c} &=\sum_{n=0}^\infty A_n (c-c)^n\\ &=\underbrace{A_0}_{n=0} +\underbrace{0}_{n=1} +\underbrace{0}_{n=2} +\underbrace{0}_{n=3} +\cdots \end{align*}

and so simply converges to $A_0\text{.}$

We now know that a power series converges when $x=c\text{.}$ We can now use our convergence tests to determine for what other values of $x$ the series converges. Perhaps most straightforward is the ratio test. The $n^{\rm th}$ term in the series $\sum_{n=0}^\infty A_n(x-c)^n$ is $a_n=A_n(x-c)^n\text{.}$ To apply the ratio test we need to compute the limit

\begin{align*} \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right| &=\lim_{n \to \infty} \left|\frac{A_{n+1}(x-c)^{n+1}}{A_n(x-c)^n}\right|\\ &=\lim_{n \to \infty} \left|\frac{A_{n+1}}{A_n}\right|\cdot |x-c|\\ &= |x-c| \cdot \lim_{n \to \infty} \left|\frac{A_{n+1}}{A_n}\right|. \end{align*}

When we do so there are several possible outcomes.

If the limit of ratios exists and is nonzero

\begin{align*} \lim_{n\rightarrow\infty}\Big|\frac{A_{n+1}}{A_n}\Big| &= A \neq 0, \end{align*}

then the ratio test says that the series $\sum_{n=0}^\infty A_n(x-c)^n$
- converges when $A \cdot |x-c| \lt 1\text{,}$ i.e. when $|x-c| \lt \frac{1}{A}\text{,}$ and
- diverges when $A \cdot |x-c| \gt 1\text{,}$ i.e. when $|x-c| \gt \frac{1}{A}\text{.}$
Because of this, when the limit exists, the quantity
Equation 3.5.2. Radius of convergence.

\begin{align*} R&=\frac{1}{A} =\bigg[\lim_{n\rightarrow\infty}\Big|\frac{A_{n+1}}{A_n}\Big|\bigg]^{-1} \end{align*}
is called the radius of convergence of the series ³.
If the limit of ratios exists and is zero
\begin{align*} \lim_{n\rightarrow\infty}\Big|\frac{A_{n+1}}{A_n}\Big| &= 0 \end{align*}
then $\lim_{n\rightarrow\infty}\Big|\frac{A_{n+1}}{A_n}\Big||x-c| =0$ for every $x$ and the ratio test tells us that the series $\sum_{n=0}^\infty A_n(x-c)^n$ converges for every number $x\text{.}$ In this case we say that the series has an infinite radius of convergence.
If the limit of ratios diverges to $+\infty$
\begin{align*} \lim_{n\rightarrow\infty}\Big|\frac{A_{n+1}}{A_n}\Big| &= +\infty \end{align*}
then $\lim_{n\rightarrow\infty}\Big|\frac{A_{n+1}}{A_n}\Big||x-c| =+\infty$ for every $x\ne c\text{.}$ The ratio test then tells us that the series $\sum_{n=0}^\infty A_n(x-c)^n$ diverges for every number $x\ne c\text{.}$ As we have seen above, when $x=c\text{,}$ the series reduces to $A_0+0+0+0+0+\cdots\text{,}$ which of course converges. In this case we say that the series has radius of convergence zero.
If $\Big|\frac{A_{n+1}}{A_n}\Big|$ does not approach a limit as $n\rightarrow\infty\text{,}$ then we learn nothing from the ratio test and we must use other tools to understand the convergence of the series.

All of these possibilities do happen. We give an example of each below. But first, the concept of “radius of convergence” is important enough to warrant a formal definition.

Definition 3.5.3.

Let $0 \lt R \lt \infty\text{.}$ If $\sum_{n=0}^\infty A_n(x-c)^n$ converges for $|x-c| \lt R\text{,}$ and diverges for $|x-c| \gt R\text{,}$ then we say that the series has radius of convergence $R\text{.}$
If $\sum_{n=0}^\infty A_n(x-c)^n$ converges for every number $x\text{,}$ we say that the series has an infinite radius of convergence.
If $\sum_{n=0}^\infty A_n(x-c)^n$ diverges for every $x\ne c\text{,}$ we say that the series has radius of convergence zero.

Example 3.5.4. Finite nonzero radius of convergence.

We already know that, if $a\ne 0\text{,}$ the geometric series $\sum\limits_{n=0}^\infty a x^n$ converges when $|x| \lt 1$ and diverges when $|x|\ge 1\text{.}$ So, in the terminology of Definition 3.5.3, the geometric series has radius of convergence $R=1\text{.}$ As a consistency check, we can also compute $R$ using 3.5.2. The series $\sum\limits_{n=0}^\infty a x^n$ has $A_n=a\text{.}$ So

\begin{equation*} R=\bigg[\lim_{n\rightarrow\infty}\Big|\frac{A_{n+1}}{A_n}\Big|\bigg]^{-1} =\Big[\lim_{n\rightarrow\infty}1\Big]^{-1} =1 \end{equation*}

as expected.

Example 3.5.5. Radius of convergence = $+\infty$.

The series $\sum\limits_{n=0}^\infty \frac{x^n}{n!}$ has $A_n=\frac{1}{n!}\text{.}$ So

\begin{align*} \lim_{n\rightarrow\infty}\Big|\frac{A_{n+1}}{A_n}\Big| &=\lim_{n\rightarrow\infty}\frac{\frac{1}{(n+1)!}}{\frac{1}{n!}}\\ & =\lim_{n\rightarrow\infty}\frac{n!}{(n+1)!}\\ & =\lim_{n\rightarrow\infty}\frac{1\times 2\times 3\times\cdots\times n} {1\times 2\times 3\times\cdots\times n\times(n+1)}\\ &=\lim_{n\rightarrow\infty}\frac{1}{n+1}\\ &=0 \end{align*}

and $\sum\limits_{n=0}^\infty \frac{x^n}{n!}$ has radius of convergence $\infty\text{.}$ It converges for every $x\text{.}$

Example 3.5.6. Radius of convergence = 0.

The series $\sum\limits_{n=0}^\infty n! x^n$ has $A_n=n!\text{.}$ So

\begin{align*} \lim_{n\rightarrow\infty}\Big|\frac{A_{n+1}}{A_n}\Big| &=\lim_{n\rightarrow\infty}\frac{(n+1)!}{n!} =\lim_{n\rightarrow\infty} \frac {1\times 2\times 3\times 4\times\cdots\times n\times(n+1)} {1\times 2\times 3\times 4\times\cdots\times n}\\ &=\lim_{n\rightarrow\infty}(n+1)\\ &=+\infty \end{align*}

and $\sum\limits_{n=0}^\infty n! x^n$ has radius of convergence zero ⁴. It converges only for $x=0\text{,}$ where it takes the value $0!=1\text{.}$

Example 3.5.7. An awkward series to test.

Comparing the series

\begin{alignat*}{7} &\ 1&+&\ 2x&+&x^2&+&2x^3&+&x^4&+&2x^5&+\cdots \\\\ \end{alignat*}

\begin{alignat*}{7} \sum_{n=1}^\infty A_nx^n=&A_0&+&A_1x&+&A_2x^2&+&A_3x^3&+&A_4x^4&+&A_5x^5&+\cdots \end{alignat*}

we see that

\begin{alignat*}{7} A_0&=1&\quad A_1&=2&\quad A_2&=1&\quad A_3&=2&\quad A_4&=1&\quad A_5&=2&\quad\cdots\\ \end{alignat*}

so that

\begin{alignat*}{7} & & \frac{A_1}{A_0}&=2&\qquad \frac{A_2}{A_1}&=\frac{1}{2}&\qquad \frac{A_3}{A_2}&=2&\qquad \frac{A_4}{A_3}&=\frac{1}{2}&\qquad \frac{A_5}{A_4}&=2&\qquad\cdots \end{alignat*}

and $\frac{A_{n+1}}{A_n}$ does not converge as $n\rightarrow\infty\text{.}$ Since the limit of the ratios does not exist, we cannot tell anything from the ratio test. Nonetheless, we can still figure out for which $x$'s our power series converges.

Because every coefficient $A_n$ is either 1 or 2, the $n^{\rm th}$ term in our series obeys
\begin{equation*} \big|A_nx^n\big|\le 2 |x|^n \end{equation*}
and so is smaller than the $n^{\rm th}$ term in the geometric series $\sum_{n=0}^\infty 2|x|^n\text{.}$ This geometric series converges if $|x| \lt 1\text{.}$ So, by the comparison test, our series converges for $|x| \lt 1$ too.
Since every $A_n$ is at least one, the $n^{\rm th}$ term in our series obeys
\begin{equation*} \big|A_nx^n\big|\ge |x|^n \end{equation*}
If $|x|\ge 1\text{,}$ this $a_n=A_n x^n$ cannot converge to zero as $n\rightarrow\infty\text{,}$ and our series diverges by the divergence test.

In conclusion, our series converges if and only if $|x| \lt 1\text{,}$ and so has radius of convergence $1\text{.}$

Example 3.5.8. A series from $\pi$.

Lets construct a series from the digits of $\pi\text{.}$ Now to avoid dividing by zero, let us set

\begin{align*} A_n &= 1 + \text{the $n^\mathrm{th}$ digit of $\pi$} \end{align*}

Since $\pi = 3.141591\dots$

\begin{gather*} A_0 = 4 \quad A_1 = 2 \quad A_2 = 5 \quad A_3 = 2 \quad A_4 = 6 \quad A_5 = 10 \quad A_6 = 2 \quad \cdots \end{gather*}

Consequently every $A_n$ is an integer between 1 and 10 and gives us the series

\begin{equation*} \sum_{n=0}^\infty A_n x^n = 4 + 2x + 5x^2 + 2x^3 + 6x^4 +10 x^5 + \cdots \end{equation*}

The number $\pi$ is irrational ⁵ and consequently the ratio $\frac{A_{n+1}}{A_n}$ cannot have a limit as $n\rightarrow\infty\text{.}$ If you do not understand why this is the case then don't worry too much about it ⁶. As in the last example, the limit of the ratios does not exist and we cannot tell anything from the ratio test. But we can still figure out for which $x$'s it converges.

Because every coefficient $A_n$ is no bigger (in magnitude) than 10, the $n^{\rm th}$ term in our series obeys
\begin{equation*} \big|A_nx^n\big|\le 10 |x|^n \end{equation*}
and so is smaller than the $n^{\rm th}$ term in the geometric series $\sum_{n=0}^\infty 10|x|^n\text{.}$ This geometric series converges if $|x| \lt 1\text{.}$ So, by the comparison test, our series converges for $|x| \lt 1$ too.
Since every $A_n$ is at least one, the $n^{\rm th}$ term in our series obeys
\begin{equation*} \big|A_nx^n\big|\ge |x|^n \end{equation*}
If $|x|\ge 1\text{,}$ this $a_n=A_n x^n$ cannot converge to zero as $n\rightarrow\infty\text{,}$ and our series diverges by the divergence test.

In conclusion, our series converges if and only if $|x| \lt 1\text{,}$ and so has radius of convergence $1\text{.}$

Though we won't prove it, it is true that every power series has a radius of convergence, whether or not the limit $\lim\limits_{n\rightarrow\infty}\Big|\frac{A_{n+1}}{A_n}\Big|$ exists.

Theorem 3.5.9.

Let $\sum\limits_{n=0}^\infty A_n (x-c)^n$ be a power series. Then one of the following alternatives must hold.

The power series converges for every number $x\text{.}$ In this case we say that the radius of convergence is $\infty\text{.}$
There is a number $0 \lt R \lt \infty$ such that the series converges for $|x-c| \lt R$ and diverges for $|x-c| \gt R\text{.}$ Then $R$ is called the radius of convergence.
The series converges for $x=c$ and diverges for all $x\ne c\text{.}$ In this case, we say that the radius of convergence is $0\text{.}$

Definition 3.5.10.

Consider the power series

\begin{gather*} \sum_{n=0}^\infty A_n (x-c)^n. \end{gather*}

The set of real $x$-values for which it converges is called the interval of convergence of the series.

Suppose that the power series $\sum\limits_{n=0}^\infty A_n (x-c)^n$ has radius of convergence $R\text{.}$ Then from Theorem 3.5.9, we have that

if $R=\infty\text{,}$ then its interval of convergence is $-\infty \lt x \lt \infty\text{,}$ which is also denoted $(-\infty,\infty)\text{,}$ and
if $R=0\text{,}$ then its interval of convergence is just the point $x=c\text{,}$ and
if $0 \lt R \lt \infty\text{,}$ then we know that the series converges for any $x$ which obeys

\begin{align*} |x-c| \lt R\quad&\text{or equivalently}\quad -R \lt x-c \lt R\\ &\text{or equivalently}\quad c-R \lt x \lt c+R \end{align*}

But we do not (yet) know whether or not the series converges at the two end points of that interval. We do know, however, that its interval of convergence must be one of
- $c-R \lt x \lt c+R\text{,}$ which is also denoted $(c-R\,,\,c+R)\text{,}$ or
- $c-R\le x \lt c+R\text{,}$ which is also denoted $[c-R\,,\,c+R)\text{,}$ or
- $c-R \lt x\le c+R\text{,}$ which is also denoted $(c-R\,,\,c+R]\text{,}$ or
- $c-R\le x \le c+R\text{,}$ which is also denoted $[c-R\,,\,c+R]\text{.}$

To reiterate — while the radius convergence, $R$ with $0 \lt R \lt \infty\text{,}$ tells us that the series converges for $|x-c| \lt R$ and diverges for $|x-c| \gt R\text{,}$ it does not (by itself) tell us whether or not the series converges when $|x-c|=R\text{,}$ i.e. when $x=c\pm R\text{.}$ The following example shows that all four possibilities can occur.

Example 3.5.11. The series $\sum_{n=1}^\infty \frac{x^n}{n^p}$.

Let $p$ be any real number⁷ and consider the series $\sum_{n=1}^\infty \frac{x^n}{n^p}\text{.}$ This series has $A_n= \frac{1}{n^p}\text{.}$ Since

\begin{equation*} \lim_{n\rightarrow\infty}\Big|\frac{A_{n+1}}{A_n}\Big| =\lim_{n\rightarrow\infty}\frac{n^p}{{(n+1)}^p} =\lim_{n\rightarrow\infty}\frac{1}{{(1+\frac{1}{n})}^p} =1 \end{equation*}

the series has radius of convergence $1\text{.}$ So it certainly converges for $|x| \lt 1$ and diverges for $|x| \gt 1\text{.}$ That just leaves $x=\pm 1\text{.}$

When $x=1\text{,}$ the series reduces to $\sum_{n=1}^\infty \frac{1}{n^p}\text{.}$ We know, from Example 3.3.6, that this series converges if and only if $p \gt 1\text{.}$
When $x=-1\text{,}$ the series reduces to $\sum_{n=1}^\infty \frac{(-1)^n}{n^p}\text{.}$ By the alternating series test, Theorem 3.3.14, this series converges whenever $p \gt 0$ (so that $\frac{1}{n^p}$ tends to zero as $n$ tends to infinity). When $p\le 0$ (so that $\frac{1}{n^p}$ does not tend to zero as $n$ tends to infinity), it diverges by the divergence test, Theorem 3.3.1.

The power series $\sum_{n=1}^\infty x^n$ (i.e. $p=0$) has interval of convergence $-1 \lt x \lt 1\text{.}$
The power series $\sum_{n=1}^\infty \frac{x^n}{n}$ (i.e. $p=1$) has interval of convergence $-1\le x \lt 1\text{.}$
The power series $\sum_{n=1}^\infty \frac{(-1)^n}{n}x^n$ (i.e. $p=1$) has interval of convergence $-1 \lt x\le 1\text{.}$
The power series $\sum_{n=1}^\infty \frac{x^n}{n^2}$ (i.e. $p=2$) has interval of convergence $-1\le x\le 1\text{.}$

Example 3.5.12. Playing with intervals of convergence.

We are told that a certain power series with centre $c=3\text{,}$ converges at $x=4$ and diverges at $x=1\text{.}$ What else can we say about the convergence or divergence of the series for other values of $x\text{?}$

We are told that the series is centred at $3\text{,}$ so its terms are all powers of $(x-3)$ and it is of the form

\begin{gather*} \sum_{n \geq 0} A_n (x-3)^n. \end{gather*}

A good way to summarise the convergence data we are given is with a figure like the one below. Green dots mark the values of $x$ where the series is known to converge. (Recall that every power series converges at its centre.) The red dot marks the value of $x$ where the series is known to diverge. The bull's eye marks the centre.

Can we say more about the convergence and/or divergence of the series for other values of $x\text{?}$ Yes!

Let us think about the radius of convergence, $R\text{,}$ of the series. We know that it must exist and the information we have been given allows us to bound $R\text{.}$ Recall that

the series converges at $x$ provided that $|x-3| \lt R$ and
the series diverges at $x$ if $|x-3| \gt R\text{.}$

We have been told that

the series converges when $x=4\text{,}$ which tells us that
- $x=4$ cannot obey $|x-3| \gt R$ so
- $x=4$ must obey $|x-3|\le R\text{,}$ i.e. $|4-3|\le R\text{,}$ i.e. $R\ge 1$
the series diverges when $x=1$ so we also know that
- $x=1$ cannot obey $|x-3| \lt R$ so
- $x=1$ must obey $|x-3|\ge R\text{,}$ i.e. $|1-3|\ge R\text{,}$ i.e. $R\le 2$

We still don't know $R$ exactly. But we do know that $1\le R\le 2\text{.}$ Consequently,

since $1$ is the smallest that $R$ could be, the series certainly converges at $x$ if $|x-3| \lt 1\text{,}$ i.e. if $2 \lt x \lt 4$ and
since $2$ is the largest that $R$ could be, the series certainly diverges at $x$ if $|x-3| \gt 2\text{,}$ i.e. if $x \gt 5$ or if $x \lt 1\text{.}$

The following figure provides a resume of all of this convergence data — there is convergence at green $x$'s and divergence at red $x$'s.

Notice that from the data given we cannot say anything about the convergence or divergence of the series on the intervals $(1,2]$ and $(4,5]\text{.}$

One lesson that we can derive from this example is that,

if a series has centre $c$ and converges at $a\text{,}$
then it also converges at all points between $c$ and $a\text{,}$ as well as at all points of distance strictly less than $|a-c|$ from $c$ on the other side of $c$ from $a\text{.}$

Subsection 3.5.2 Working With Power Series

Just as we have done previously with limits, differentiation and integration, we can construct power series representations of more complicated functions by using those of simpler functions. Here is a theorem that helps us to do so.

Theorem 3.5.13. Operations on Power Series.

Assume that the functions $f(x)$ and $g(x)$ are given by the power series

\begin{equation*} f(x) = \sum_{n=0}^\infty A_n (x-c)^n \qquad g(x) = \sum_{n=0}^\infty B_n (x-c)^n \end{equation*}

for all $x$ obeying $|x-c| \lt R\text{.}$ In particular, we are assuming that both power series have radius of convergence at least $R\text{.}$ Also let $K$ be a constant. Then

\begin{align*} f(x)+g(x) &= \sum_{n=0}^\infty [A_n+B_n]\, (x-c)^n\\ Kf(x) &= \sum_{n=0}^\infty K\, A_n\, (x-c)^n\\ (x-c)^Nf(x) &= \sum_{n=0}^\infty A_n\, (x-c)^{n+N} \quad\text{for any integer $N\ge 1$}\\ &= \sum_{k=N}^\infty A_{k-N}\, (x-c)^k \quad\text{where $k=n+N$}\\ f'(x) &= \sum_{n=0}^\infty A_n\, n\,(x-c)^{n-1} = \sum_{n=1}^\infty A_n\, n\,(x-c)^{n-1}\\ \int_c^x f(t)\dee{t} &= \sum_{n=0}^\infty A_n \frac{(x-c)^{n+1}}{n+1}\\ \int f(x)\ \dee{x} &= \bigg[\sum_{n=0}^\infty A_n \frac{(x-c)^{n+1}}{n+1}\bigg]+C \quad\text{with $C$ an arbitrary constant} \end{align*}

for all $x$ obeying $|x-c| \lt R\text{.}$

In particular the radius of convergence of each of the six power series on the right hand sides is at least $R\text{.}$ In fact, if $R$ is the radius of convergence of $\sum\limits_{n=0}^\infty A_n (x-c)^n\text{,}$ then $R$ is also the radius of convergence of all of the above right hand sides, with the possible exceptions of $\sum\limits_{n=0}^\infty [A_n+B_n]\, (x-c)^n$ and $\sum\limits_{n=0}^\infty KA_n\, (x-c)^n$ when $K=0\text{.}$

Example 3.5.14. More on the last part of Theorem 3.5.13.

The last statement of Theorem 3.5.13 might seem a little odd, but consider the following two power series centred at $0\text{:}$

\begin{align*} \sum_{n=0}^\infty 2^n x^n & \text{ and } \sum_{n=0}^\infty (1-2^n) x^n. \end{align*}

The ratio test tells us that they both have radius of convergence $R=\frac{1}{2}\text{.}$ However their sum is

\begin{align*} \sum_{n=0}^\infty 2^n x^n + \sum_{n=0}^\infty (1-2^n) x^n &= \sum_{n=0}^\infty x^n \end{align*}

which has the larger radius of convergence $1\text{.}$

A more extreme example of the same phenomenon is supplied by the two series

\begin{align*} \sum_{n=0}^\infty 2^n x^n & \text{ and } \sum_{n=0}^\infty (-2^n) x^n. \end{align*}

They are both geometric series with radius of convergence $R=\frac{1}{2}\text{.}$ But their sum is

\begin{align*} \sum_{n=0}^\infty 2^n x^n + \sum_{n=0}^\infty (-2^n) x^n &= \sum_{n=0}^\infty (0)x^n \end{align*}

which has radius of convergence $+\infty\text{.}$

We'll now use this theorem to build power series representations for a bunch of functions out of the one simple power series representation that we know — the geometric series

\begin{equation*} \frac{1}{1-x} = \sum_{n=0}^\infty x^n\qquad \text{for all $|x| \lt 1$} \end{equation*}

Example 3.5.15. $\frac{1}{1-x^2}$.

Find a power series representation for $\frac{1}{1-x^2}\text{.}$

Solution: The secret to finding power series representations for a good many functions is to manipulate them into a form in which $\frac{1}{1-y}$ appears and use the geometric series representation $\frac{1}{1-y} = \sum_{n=0}^\infty y^n\text{.}$ We have deliberately renamed the variable to $y$ here — it does not have to be $x\text{.}$ We can use that strategy to find a power series expansion for $\frac{1}{1-x^2}$ — we just have to recognize that $\frac{1}{1-x^2}$ is the same as $\frac{1}{1-y}$ if we set $y$ to $x^2\text{.}$

\begin{align*} \frac{1}{1-x^2} &= \frac{1}{1-y}\bigg|_{y=x^2} =\bigg[\sum_{n=0}^\infty y^n \bigg]_{y=x^2} \quad\text{if $|y| \lt 1$, i.e. $|x| \lt 1$}\\ &= \sum_{n=0}^\infty {\big(x^2\big)}^n = \sum_{n=0}^\infty x^{2n}\\ &= 1+x^2+x^4+x^6 +\cdots\hskip-0.2in \end{align*}

This is a perfectly good power series. There is nothing wrong with the power of $x$ being $2n\text{.}$ (This just means that the coefficients of all odd powers of $x$ are zero.) In fact, you should try to always write power series in forms that are as easy to understand as possible. The geometric series that we used at the end of the first line converges for

\begin{gather*} |y| \lt 1 \iff \big|x^2\big| \lt 1 \iff |x| \lt 1 \end{gather*}

So our power series has radius of convergence $1$ and interval of convergence $-1 \lt x \lt 1\text{.}$

Example 3.5.16. $\frac{x}{2+x^2}$.

Find a power series representation for $\frac{x}{2+x^2}\text{.}$

Solution: This example is just a more algebraically involved variant of the last one. Again, the strategy is to manipulate $\frac{x}{2+x^2}$ into a form in which $\frac{1}{1-y}$ appears.

\begin{align*} \frac{x}{2+x^2} &= \frac{x}{2}\ \frac{1}{1+\frac{x^2}{2}} =\ \frac{x}{2}\ \frac{1}{1-\big({-}\frac{x^2}{2}\big)} \qquad \text{set $-\frac{x^2}{2}=y$}\\ &= \frac{x}{2}\ \frac{1}{1-y}\bigg|_{y=-\frac{x^2}{2}} = \frac{x}{2}\ \bigg[\sum_{n=0}^\infty y^n \bigg]_{y=-\frac{x^2}{2}} \quad\text{if $|y| \lt 1$}\\ \end{align*}

Now use Theorem 3.5.13 twice

\begin{align*} &= \frac{x}{2}\ \sum_{n=0}^\infty {\Big(-\frac{x^2}{2}\Big)}^n =\ \frac{x}{2}\ \sum_{n=0}^\infty \frac{(-1)^n}{2^n}x^{2n} = \sum_{n=0}^\infty \frac{(-1)^n}{2^{n+1}}x^{2n+1}\\ &= \frac{x}{2} - \frac{x^3}{4} + \frac{x^5}{8}-\frac{x^7}{16} +\cdots \end{align*}

The geometric series that we used in the second line converges when

\begin{align*} |y| \lt 1 &\iff \big|{-}\frac{x^2}{2}\big| \lt 1\\ & \iff |x|^2 \lt 2 \iff |x| \lt \sqrt{2} \end{align*}

So the given power series has radius of convergence $\sqrt{2}$ and interval of convergence $-\sqrt{2} \lt x \lt \sqrt{2}\text{.}$

Example 3.5.17. Nonzero centre.

Find a power series representation for $\frac{1}{5-x}$ with centre $3\text{.}$

Solution: The new wrinkle in this example is the requirement that the centre be $3\text{.}$ That the centre is to be $3$ means that we need a power series in powers of $x-c\text{,}$ with $c=3\text{.}$ So we are looking for a power series of the form $\sum_{n=0}^\infty A_n(x-3)^n\text{.}$ The easy way to find such a series is to force an $x-3$ to appear by adding and subtracting a $3\text{.}$

\begin{gather*} \frac{1}{5-x} =\frac{1}{5-(x-3)-3} =\frac{1}{2-(x-3)} \end{gather*}

Now we continue, as in the last example, by manipulating $\frac{1}{2-(x-3)}$ into a form in which $\frac{1}{1-y}$ appears.

\begin{align*} \frac{1}{5-x} =\frac{1}{2-(x-3)} &= \frac{1}{2}\ \frac{1}{1-\frac{x-3}{2}} \qquad \text{set $\frac{x-3}{2}=y$}\\ &= \frac{1}{2}\ \frac{1}{1-y}\bigg|_{y=\frac{x-3}{2}} = \frac{1}{2}\ \bigg[\sum_{n=0}^\infty y^n \bigg]_{y=\frac{x-3}{2}} \quad\text{if $|y| \lt 1$}\\ &= \frac{1}{2}\ \sum_{n=0}^\infty {\Big(\frac{x-3}{2}\Big)}^n = \sum_{n=0}^\infty \frac{(x-3)^n}{2^{n+1}}\\ &= \frac{1}{2} + \frac{x-3}{4} + \frac{(x-3)^2}{8} + \frac{(x-3)^3}{16} +\cdots \end{align*}

The geometric series that we used in the second line converges when

\begin{align*} |y| \lt 1 & \iff \Big|\frac{x-3}{2}\Big| \lt 1\\ & \iff |x-3| \lt 2\\ & \iff -2 \lt x-3 \lt 2\\ & \iff 1 \lt x \lt 5 \end{align*}

So the power series has radius of convergence $2$ and interval of convergence $1 \lt x \lt 5\text{.}$

In the previous two examples, to construct a new series from an existing series, we replaced $x$ by a simple function. The following theorem gives us some more (but certainly not all) commonly used substitutions.

Theorem 3.5.18. Substituting in a Power Series.

Assume that the function $f(x)$ is given by the power series

\begin{equation*} f(x) = \sum_{n=0}^\infty A_n x^n \end{equation*}

for all $x$ in the interval $I\text{.}$ Also let $K$ and $k$ be real constants. Then

\begin{align*} f\big(Kx^k\big) &= \sum_{n=0}^\infty A_nK^n\, x^{kn} \end{align*}

whenever $Kx^k$ is in $I\text{.}$ In particular, if $\sum_{n=0}^\infty A_n x^n$ has radius of convergence $R\text{,}$ $K$ is nonzero and $k$ is a natural number, then $\sum_{n=0}^\infty A_nK^n\, x^{kn}$ has radius of convergence $\root{k}\of{R/|K|}\text{.}$

Example 3.5.19. $\frac{1}{(1-x)^2}$.

Find a power series representation for $\frac{1}{(1-x)^2}\text{.}$

Solution: Once again the trick is to express $\frac{1}{(1-x)^2}$ in terms of $\frac{1}{1-x}\text{.}$ Notice that

\begin{align*} \frac{1}{(1-x)^2} &= \diff{}{x}\left\{\frac{1}{1-x}\right\}\\ &= \diff{}{x} \left\{\sum_{n=0}^\infty x^n \right\}\\ &= \sum_{n=1}^\infty nx^{n-1} \qquad\text{by Theorem }\knowl{./knowl/thm_SRpsops.html}{\text{3.5.13}} \end{align*}

Note that the $n=0$ term has disappeared because, for $n=0\text{,}$

\begin{equation*} \diff{}{x} x^n = \diff{}{x} x^0 = \diff{}{x} 1= 0 \end{equation*}

Also note that the radius of convergence of this series is one. We can see this via Theorem 3.5.13. That theorem tells us that the radius of convergence of a power series is not changed by differentiation — and since $\sum_{n=0}^\infty x^n$ has radius of convergence one, so too does its derivative.

Without much more work we can determine the interval of convergence by testing at $x=\pm 1\text{.}$ When $x=\pm 1$ the terms of the series do not go to zero as $n \to \infty$ and so, by the divergence test, the series does not converge there. Hence the interval of convergence for the series is $-1 \lt x \lt 1\text{.}$

Notice that, in this last example, we differentiated a known series to get to our answer. As per Theorem 3.5.13, the radius of convergence didn't change. In addition, in this particular example, the interval of convergence didn't change. This is not always the case. Differentiation of some series causes the interval of convergence to shrink. In particular the differentiated series may no longer be convergent at the end points of the interval ⁸. Similarly, when we integrate a power series the radius of convergence is unchanged, but the interval of convergence may expand to include one or both ends, as illustrated by the next example.

Example 3.5.20. $\log (1+x)$.

Find a power series representation for $\log (1+x)\text{.}$

Solution: Recall that $\diff{}{x}\log (1+x) = \frac{1}{1+x}$ so that $\log(1+t)$ is an antiderivative of $\frac{1}{1+t}$ and

\begin{align*} \log(1+x)\ &=\ \int_0^x \frac{\dee{t}}{1+t} \ =\ \int_0^x\Big[\sum_{n=0}^\infty (-t)^n\Big]\dee{t}\\ &=\ \sum_{n=0}^\infty \int_0^x (-t)^n\dee{t} \qquad\text{by Theorem }\knowl{./knowl/thm_SRpsops.html}{\text{3.5.13}}\\ &=\ \sum_{n=0}^\infty (-1)^n\frac{x^{n+1}}{n+1}\\ &=\ x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots \end{align*}

Theorem 3.5.13 guarantees that the radius of convergence is exactly one (the radius of convergence of the geometric series $\sum_{n=0}^\infty (-t)^n$) and that

\begin{gather*} \log(1+x) = \sum_{n=0}^\infty (-1)^n\frac{x^{n+1}}{n+1} \qquad\text{for all}\quad -1 \lt x \lt 1 \end{gather*}

When $x=-1$ our series reduces to $\sum_{n=0}^\infty \frac{-1}{n+1}\text{,}$ which is (minus) the harmonic series and so diverges. That's no surprise — $\log(1+(-1)) =\log 0=-\infty\text{.}$ When $x=1\text{,}$ the series converges by the alternating series test. It is possible to prove, by continuity, though we won't do so here, that the sum is $\log 2\text{.}$ So the interval of convergence is $-1 \lt x\le 1\text{.}$

Example 3.5.21. $\arctan x$.

Find a power series representation for $\arctan x\text{.}$

Solution: Recall that $\diff{}{x}\arctan x = \frac{1}{1+x^2}$ so that $\arctan t$ is an antiderivative of $\frac{1}{1+t^2}$ and

\begin{align*} \arctan x\ &=\ \int_0^x \frac{\dee{t}}{1+t^2} \ =\ \int_0^x\Big[\sum_{n=0}^\infty {(-t^2)}^n\Big]\dee{t} \ =\ \sum_{n=0}^\infty \int_0^x (-1)^n t^{2n}\dee{t}\\ &=\ \sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{2n+1}\\ &=x -\frac{x^3}{3} +\frac{x^5}{5}-\cdots \end{align*}

Theorem 3.5.13 guarantees that the radius of convergence is exactly one (the radius of convergence of the geometric series $\sum_{n=0}^\infty (-t^2)^n$) and that

\begin{align*} \arctan x\ &=\ \sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{2n+1} \qquad\text{for all $-1 \lt x \lt 1$} \end{align*}

When $x=\pm 1\text{,}$ the series converges by the alternating series test. So the interval of convergence is $-1\le x\le 1\text{.}$ It is possible to prove, though once again we won't do so here, that when $x=\pm 1\text{,}$ the series $\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{2n+1}$ converges to the value of the left hand side, $\arctan x\text{,}$ at $x=\pm 1\text{.}$ That is, to $\arctan(\pm 1)=\pm \frac{\pi}{4}\text{.}$

The operations on power series dealt with in Theorem 3.5.13 are fairly easy to apply. Unfortunately taking the product, ratio or composition of two power series is more involved and is beyond the scope of this course ⁹. Unfortunately Theorem 3.5.13 alone will not get us power series representations of many of our standard functions (like $e^x$ and $\sin x$). Fortunately we can find such representations by extending Taylor polynomials ¹⁰ to Taylor series.

Exercises 3.5.3 Exercises

Exercises — Stage 1 .

1.

Suppose $f(x)=\displaystyle\sum_{n=0}^\infty\left(\dfrac{3-x}{4}\right)^n\text{.}$ What is $f(1)\text{?}$

2.

Suppose $f(x)=\displaystyle\sum_{n=1}^\infty \dfrac{(x-5)^n}{n!+2}\text{.}$ Give a power series representation of $f'(x)\text{.}$

3.

Let $f(x)=\displaystyle \sum_{n=a}^\infty A_n(x-c)^n$ for some positive constants $a$ and $c\text{,}$ and some sequence of constants $\{A_n\}\text{.}$ For which values of $x$ does $f(x)$ definitely converge?

4.

Let $f(x)$ be a power series centred at $c=5\text{.}$ If $f(x)$ converges at $x=-1\text{,}$ and diverges at $x=11\text{,}$ what is the radius of convergence of $f(x)\text{?}$

Exercises — Stage 2 .

5. (✳).

(a) Find the radius of convergence of the series

\begin{equation*} \sum_{k=0}^\infty (-1)^k 2^{k+1} x^k \end{equation*}

(b) You are given the formula for the sum of a geometric series, namely:

\begin{equation*} 1+r+r^2 + \cdots =\frac{1}{1-r},\qquad|r| \lt 1 \end{equation*}

Use this fact to evaluate the series in part (a).

6. (✳).

Find the radius of convergence for the power series $\displaystyle\sum_{k=0}^\infty \frac{x^k}{10^{k+1}(k+1)!}$

7. (✳).

Find the radius of convergence for the power series $\displaystyle\sum_{n=0}^\infty \frac{(x - 2)^n}{n^2+1}\text{.}$

8. (✳).

Consider the power series $\displaystyle\sum\limits_{n=1}^\infty \frac{(-1)^n(x+2)^n}{\sqrt{n}}\text{,}$ where $x$ is a real number. Find the interval of convergence of this series.

9. (✳).

Find the radius of convergence and interval of convergence of the series

\begin{gather*} \sum_{n=0}^{\infty} \frac{(-1)^n}{n+1} \left(\frac{x+1}{3}\right)^n \end{gather*}

10. (✳).

Find the interval of convergence for the power series

\begin{gather*} \sum_{n=1}^\infty \frac{(x-2)^n}{n^{4/5}(5^n-4)}. \end{gather*}

11. (✳).

Find all values $x$ for which the series $\displaystyle\sum_{n=1}^\infty\frac{(x+2)^n}{n^2}$ converges.

12. (✳).

Find the interval of convergence for $\displaystyle \sum_{n=1}^\infty \frac{4^n}{n}(x-1)^n\text{.}$

13. (✳).

Find, with explanation, the radius of convergence and the interval of convergence of the power series

\begin{equation*} \sum_{n=0}^\infty (-1)^n\frac{(x-1)^n}{2^n(n+2)} \end{equation*}

14. (✳).

Find the interval of convergence for the series $\displaystyle \sum_{n=1}^\infty (-1)^n n^2(x-a)^{2n}$ where $a$ is a constant.

15. (✳).

Find the interval of convergence of the following series:

${\displaystyle \sum_{k=1}^\infty \frac{(x+1)^k}{k^2 9^k}}\text{.}$
${\displaystyle \sum_{k=1}^\infty a_k(x-1)^k}\text{,}$ where $a_k \gt 0$ for $k=1,2,\cdots$ and $\ {\displaystyle \sum_{k=1}^\infty \Big(\frac{a_k}{a_{k+1}} -\frac{a_{k+1}}{a_{k+2}}\Big) =\frac{a_1}{a_2} }\text{.}$

16. (✳).

Find a power series representation for $\dfrac{x^3}{1-x}\text{.}$

17.

Suppose $f'(x)=\displaystyle\sum_{n=0}^\infty \frac{(x-1)^{n}}{n+2}\text{,}$ and $\displaystyle \int_5^x f(t)\dee{t}=3x+\displaystyle\sum_{n=1}^\infty \frac{(x-1)^{n+1}}{n(n+1)^2}\text{.}$

Give a power series representation of $f(x)\text{.}$

Exercises — Stage 3 .

18. (✳).

Determine the values of $x$ for which the series

\begin{equation*} \sum_{n=2}^\infty\frac{x^n}{3^{2n}\log n} \end{equation*}

converges absolutely, converges conditionally, or diverges.

19. (✳).

(a) Find the power--series representation for $\displaystyle\int\frac{1}{1+x^3}\,\dee{x}$ centred at $0$ (i.e. in powers of $x$).

(b) The power series above is used to approximate $\displaystyle\int_0^{1/4}\frac{1}{1+x^3}\,\dee{x}\text{.}$ How many terms are required to guarantee that the resulting approximation is within $10^{-5}$ of the exact value? Justify your answer.

20. (✳).

(a) Show that $\displaystyle\sum_{n=0}^\infty nx^n =\frac{x}{(1-x)^2}$ for $-1 \lt x \lt 1\text{.}$

(b) Express $\displaystyle\sum_{n=0}^\infty n^2x^n$ as a ratio of polynomials. For which $x$ does this series converge?

21. (✳).

Suppose that you have a sequence $\{b_n\}$ such that the series $\sum_{n=0}^{\infty}(1-b_n)$ converges. Using the tests we've learned in class, prove that the radius of convergence of the power series $\displaystyle\sum_{n=0}^{\infty}b_nx^n$ is equal to $1\text{.}$

22. (✳).

Assume $\big\{a_n \big\}$ is a sequence such that $na_n$ decreases to $C$ as $n \rightarrow\infty$ for some real number $C \gt 0$

(a) Find the radius of convergence of $\displaystyle\sum\limits_{n=1}^\infty a_n x^n$ . Justify your answer carefully.

(b) Find the interval of convergence of the above power series, that is, find all $x$ for which the power series in (a) converges. Justify your answer carefully.

23.

An infinitely long, straight rod of negligible mass has the following weights:

At every whole number $n\text{,}$ a mass of weight $\dfrac{1}{2^n}$ at position $n\text{,}$ and
a mass of weight $\dfrac{1}{3^n}$ at position $-n\text{.}$

At what position is the centre of mass of the rod?

24.

Let $f(x)=\displaystyle\sum_{n=0}^{\infty}A_n(x-c)^n\text{,}$ for some constant $c$ and a sequence of constants $\{A_n\}\text{.}$ Further, let $f(x)$ have a positive radius of covergence.

If $A_1=0\text{,}$ show that $y=f(x)$ has a critical point at $x=c\text{.}$ What is the relationship between the behaviour of the graph at that point and the value of $A_2\text{?}$

25.

Evaluate $\displaystyle \sum_{n=3}^\infty \frac{n}{5^{n-1}}\text{.}$

26.

Find a polynomial that approximates $f(x)=\log(1+ x)$ to within an error of $10^{-5}$ for all values of $x$ in $\left(0,\frac{1}{10}\right)\text{.}$

Then, use your polynomial to approximate $\log(1.05)$ as a rational number.

27.

Find a polynomial that approximates $f(x)=\arctan x$ to within an error of $10^{-5}$ for all values of $x$ in $\left(-\frac{1}{4},\frac{1}{4}\right)\text{.}$

Recall that $n!=1\times 2\times 3\times\cdots\times n$ is called “$n$ factorial”. By convention $0!=1\text{.}$

By convention, when the term $(x-c)^0$ appears in a power series, it has value $1$ for all values of $x\text{,}$ even $x=c\text{.}$

The use of the word “radius” might seem a little odd here, since we are really describing the interval in the real line where the series converges. However, when one starts to consider power series over complex numbers, the radius of convergence does describe a circle inside the complex plane and so “radius” is a more natural descriptor.

Because of this, it might seem that such a series is fairly pointless. However there are all sorts of mathematical games that can be played with them without worrying about their convergence. Such “formal” power series can still impart useful information and the interested reader is invited to look up “generating functions” with their preferred search engine.

We give a proof of this in the optional §3.7 at the end of this chapter.

This is a little beyond the scope of the course. Roughly speaking, think about what would happen if the limit of the ratios did exist. If the limit were smaller than $1\text{,}$ then it would tell you that the terms of our series must be getting smaller and smaller and smaller — which is impossible because they are all integers between 1 and 10. Similarly if the limit existed and were bigger than $1$ then the terms of the series would have to get bigger and bigger and bigger — also impossible. Hence if the ratio exists then it must be equal to 1 — but in that case because the terms are integers, they would have to be all equal when $n$ became big enough. But that means that the expansion of $\pi$ would be eventually periodic — something that only rational numbers do (a proof is given in the optional §3.7 at the end of this chapter).

We avoid problems with $0^p$ by starting the series from $n=1\text{.}$

Consider the power series $\sum_{n=1}^\infty \frac{x^n}{n}\text{.}$ We know that its interval of convergence is $-1 \leq x \lt 1\text{.}$ (Indeed see the next example.) When we differentiate the series we get the geometric series $\sum_{n=0}^\infty x^n$ which has interval of convergence $-1 \lt x \lt 1\text{.}$

As always, a quick visit to your favourite search engine will direct the interested reader to more information.

Now is a good time to review your notes from last term, though we'll give you a whirlwind review over the next page or two.