# Infinite Summation

#### bincybn

##### Member
Hii All,

Can anyone give me a hint to evaluate $$\displaystyle \sum_{n=1}^{\infty}\frac{a^{n}}{n^{1-m}}$$; Here $$\displaystyle 0<m,\,a<1$$.

Please note that the summation converges and $$\displaystyle < \frac{a}{1-a}$$.

A tighter upper bound can be achieved as $$\displaystyle 1+\int_{1}^{\infty}\frac{a^{x}}{x^{1-m}}dx$$.

Is there any way to get the exact summation?

Thanks and regards,

Bincy

Last edited by a moderator:

#### Sudharaka

##### Well-known member
MHB Math Helper
Hii All,

Can anyone give me a hint to evaluate $$\displaystyle \sum_{n=1}^{\infty}\frac{a^{n}}{n^{1-m}}$$; Here $$\displaystyle 0<m,\,a<1$$.

Please note that the summation converges and $$\displaystyle < \frac{a}{1-a}$$.

A tighter upper bound can be achieved as $$\displaystyle 1+\int_{1}^{\infty}\frac{a^{x}}{x^{1-m}}dx$$.

Is there any way to get the exact summation?

Thanks and regards,

Bincy
Hi Bincy, This summation could be given in terms of the Polylogarithm function.

$\sum_{n=1}^{\infty}\frac{a^{n}}{n^{1-m}}=\mbox{Li}_{1-m}(a)\mbox{ for }|a|<1$