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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 \(\displaystyle 1+\int_{1}^{\infty}\frac{a^{x}}{x^{1-m}}dx\).

Thanks and regards,

Bincy

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

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