Summation: Evaluate \sum_{n=1}^{\infty}\frac{a^{n}}{n^{1-m}}

[bincy]

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

 

[Sudharaka]

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.

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