Calculate sum of series ‎$‎\sum_{k=1}^\infty (k+x-m)^\alpha e^{\beta (k+x-m)}$‎.

Question

‎I've been stuck with calculating the sum of series of the following problem. Can you help me?‎

‎ $‎\sum_{k=1}^\infty‎(k+x-m)^\alpha ‎e^{‎\beta‎(k+x-m)}‎$

for ‎$‎‎\alpha‎>0‎$‎, ‎$‎‎\beta‎<0‎$‎, ‎$‎m\in‎\mathbb{N}‎$‎ and ‎‎$‎x‎\geq ‎0‎$‎‎.

‎ I know that $‎\sum_{k=1}^\infty ‎k^\alpha ‎e^{‎\beta ‎k} = Li_{-‎\alpha}‎ (e^\beta)‎$, ‎which ‎‎$‎Li_n ‎(x)‎$ ‎is ‎the ‎‎Polylogarithm function.

Answer 1

$‎S=\sum\limits_{k=1}^{m-1} ‎(-k)^\alpha ‎e^{-‎\beta ‎k}=\sum\limits_{k=0}^{\infty} ‎(-k)^\alpha ‎e^{-‎\beta ‎k}-\sum\limits_{k=m}^{\infty} ‎(-k)^\alpha ‎e^{-‎\beta ‎k}‎$

Reindexing the second sum:

$\sum\limits_{k=0}^{\infty} ‎(-k)^\alpha ‎e^{-‎\beta ‎k}-\sum\limits_{k=0}^{\infty} ‎(-k-m)^\alpha ‎e^{-‎\beta (‎k+m)}$

S can be expressed by Lerch zeta function:

$S=(-1)^\alpha \big[L(\frac{-\beta}{2\pi i},0,-\alpha)-e^{-\beta m} L(\frac{-\beta}{2\pi i},m,-\alpha)\big]$

By the definition of polylogarthm function we have that:

$L(\frac{-\beta}{2\pi i},0,-\alpha)=\sum\limits_{k=1}^{\infty} ‎k^\alpha ‎e^{-‎\beta ‎k}=Li_{-\alpha}(e^{-\beta})$

Finally:

$S=(-1)^\alpha \big[Li_{-\alpha}(e^{-\beta})-e^{-\beta m} L(\frac{-\beta}{2\pi i},m,-\alpha)\big]$

Answer 2

The sum can be expressed by Lerch zeta function:

‎$-\dfrac{e^{\beta (x-m)}}{(x-m)^{-\alpha}}+e^{\beta (x-m)}\sum\limits_{k=0}^\infty \dfrac{e^{\beta k}}{(k+x-m)^{-\alpha}}=e^{\beta (x-m)}L(\frac{\beta}{2\pi i},x-m,-\alpha)-\dfrac{e^{\beta (x-m)}}{(x-m)^{-\alpha}}$