Series

To sum a series , place the insertion point in the series and choose Evaluate.

$\blacktriangleright$ Evaluate

$\dsum\limits_{{n=1}}^{{\infty }% }$${\dfrac{{1}}{{n^{2}}}}$ = ${\frac{{1}}{{6}}}$π²

$\dsum\limits_{{n=1}}^{{\infty }}$${\dfrac{{(-1)^{n}}}{{n}}}$ = - ln 2

$\dsum\limits_{{n=1}}^{{\infty }}$${\dfrac{{1}}{{n^{3}}}}$ = ζ$\left(\vphantom{ 3}\right.$3$\left.\vphantom{ 3}\right)$

$\dsum\limits_{{n=1}}^{{\infty }}$${\dfrac{{20^{n}}}{{n!}}}$ = e²⁰ - 1

$\dsum\limits_{{n=1}}^{{\infty }}$$\left(\vphantom{ 0.99}\right.$0.99$\left.\vphantom{ 0.99}\right)^{{n}}_{}$ = 99.0

$\dsum\limits_{{n=1}}^{{\infty }}$sin nπ = $\dsum\limits_{{n=1}}^{{\infty }}$sin nπ

$\dsum\limits_{{n=1}}^{{\infty }}$${\dfrac{{n+1}}{{n}}}$ = ∞

$\dsum\limits_{{n=1}}^{{\infty }}$${\dfrac{{n^{4}-1}}{{n^{4}}}}$ = ∞

$\dsum\limits_{{n=1}}^{{\infty }}$${\dfrac{{n^{2}-1}}{{n^{4}}}}$ = - ${\dfrac{{1}}{{90}}}$π⁴ + ${\dfrac{{1}}{{6}}}$π²

Occasionally, a result is obtained that may be obscure, such as the responses to $\dsum\limits_{{n=1}}^{{\infty }}$sin nπ and $\dsum\limits_{{n=1}}^{{\infty }}$${\frac{{1}}{{n^{3}}}}$. The lack of response to $\dsum\limits_{{n=1}}^{{\infty }}$sin nπ is the Maple version of ``I give up'' or ``I do not have enough information.'' The series $\dsum\limits_{{n=1}}^{{\infty }}$${\frac{{1}}{{n^{3}}}}$ and the values of the zeta function ζ$\left(\vphantom{ \cdot }\right.$⋅$\left.\vphantom{ \cdot }\right)$ can be estimated numerically.

$\blacktriangleright$ Evaluate Numerically

ζ$\left(\vphantom{ 3}\right.$3$\left.\vphantom{ 3}\right)$ = 1.202056903

$\dsum\limits_{{n=1}}^{{\infty }% }$${\dfrac{{1}}{{n^{3}}}}$ = 1.202056903

$\blacktriangleright$ Evaluate, Evaluate Numerically

$\dsum\limits_{{n=1}}^{{\infty }}$${\dfrac{{1}}{{n^{5}}}}$ = ζ$\left(\vphantom{ 5}\right.$5$\left.\vphantom{ 5}\right)$ = 1.036927755

To sum a series of a form similar to $\sum_{{n=1}}^{{\infty }}$a_n, first enter an equation such as a_n = ${\frac{{n^{2}}}{{2^{n}}}}$. Then, from the Define submenu, choose New Definition.