Let $m$ and $n$ are positive integers, then find the sum of the infinite series defined as $$\sum_{k=0}^\infty \frac{(-1)^k\Gamma(m+k) }{\Gamma(m)k!(k+m)^n}.$$ I was managed to the sum with $m=2$ and $3$ and whne $m=2$ the sum is $\Gamma(n)(\eta(n)-\eta(n-1)).$
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$\begingroup$ If $F_n(x):=\sum_{k=0}^\infty \frac{(x)^k\Gamma(m+k) }{\Gamma(m)k!(k+m)^n}$, then consider $\frac{d}{dx}(F_n(x) x^{k+m})$, related to $F_{n-1}(x)$. Of course $F_0(x)$ is easy. $\endgroup$– Gerald EdgarCommented Aug 31 at 19:30
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1$\begingroup$ Expression for $\frac{1}{(k+m)^n}$: $$ \frac{1}{(k+m)^n} = \frac{1}{m^n} \left( \frac{(m)_k \times (m)_k \times \cdots \times (m)_k}{(m+1)_k \times (m+1)_k \times \cdots \times (m+1)_k} \right), $$ where the products $(m)_k$ and $(m+1)_k$ are repeated $n$ times in the numerator and the denominator, respectively. Series Representation in Terms of Hypergeometric Function: $$ \sum_{k=0}^{\infty} \frac{(m)_k^{n+1} \, (-1)^k}{(m+1)_k^n \, k!} = \frac{\Gamma(n)}{m^n}{}_{n+1}F_{n}\left(m, m, \ldots, m; m+1, m+1, \ldots, m+1; -1\right). $$ $\endgroup$– user90533Commented Sep 1 at 11:43
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