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Function: csum - check convergence of a series

Calling sequence:

csum(e,n);

Parameters:

e - an expression (the summand), depending on n

n - a name

Description:

This function attempts to check whether the series

converges (where

is assumed to be sufficiently large that the terms are all defined). It returns

if it determines that the series does converge,

if it determines that the series diverges, and

if it can't show either convergence or divergence.

Note that sum is not suitable for determining convergence, as it often returns a finite result for a divergent series (which may be the result of a summability method). To restrict attention to convergent series, use csum to test for convergence before trying sum .

If infolevel[csum] is set to a positive value, csum will provide more information: whether the series converges absolutely, what were the results of each convergence test, etc.

depends on a parameter, and the convergence of the series depends on the value of the parameter, csum should return

. You can use assume to restrict the values of the parameter.

However, like many Maple procedures, csum is subject to the "specialization problem": if there are a discrete set of special values of a parameter for which series converges, while it diverges otherwise (or vice versa), csum will probably ignore the special values and return the "generic" result.

This function is part of the Maple Advisor Database library.

Examples:

A typical convergent series.

> csum(1/(n^2+1),n);

A typical divergent series.

> csum((n^2+1)/(n^3+1),n);

A series that converges conditionally. Setting infolevel[csum] gives us extra information.

> infolevel[csum]:= 1;
csum(exp(I*n)/ln(n),n);

csum:   Checking sum of   exp(I*n)/ln(n)   over   n

csum:   Series does not converge absolutely

csum/condit:   Series converges

A series whose convergence is an open question (involving how well can be represented by rational numbers). Not surprisingly, csum can't decide it.

> infolevel[csum]:= 0:
csum(1/(n^4*sin(n)^2),n);

Bugs in limit can affect csum . In this case the bug produces an incorrect result in the ratio test. This one should diverge.

> infolevel[csum]:= 2:
csum(sin(n)/2^n+1/n,n);

csum:   Checking sum of   sin(n)/(2^n)+1/n   over   n

csum/limitzero:   Checking limit of terms

csum/limitzero:   OK, limit is 0

csum/ratiotest:   Trying ratio test

csum/ratiotest:   Ratio test succeeds with ratio limit    0

csum:   Series   sin(n)/(2^n)+1/n   converges absolutely

> limit((sin(n+1)/2^(n+1)+1/(n+1))/(sin(n)/2^n+1/n),n=infinity);

A series depending on a parameter . Convergence depends on the value of .

> csum(p^n/n,n);

This one converges for all .

> csum(p^n/n!,n);

With an assumption:

> assume(p>-1,p<1): csum(p^n/n,n);

This one diverges for , but csum does not catch that.

> csum(1/(n+p*n^2),n);

See also:

assume , limit , sum

Maple Advisor Database R. Israel 2000