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I want to know if there exists any result about the exact or an approximate value of the following sum of the infinite series involving Riemann Zeta functions. Any pointer towards related results will be helpful. Thanks.

$ \sum_{n=1}^\infty \frac{(-b)^n}{n\,!} \frac{\zeta(a(n+1))}{\zeta(a)}, ~\text{ where $a, b$ are positive real constants} $

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Assuming $a>\frac12$, \begin{align*} \sum_{n=1}^\infty \frac{(-b)^n}{n!} \frac{\zeta(a(n+1))}{\zeta(a)} &= \frac1{\zeta(a)} \sum_{n=1}^\infty \frac{(-b)^n}{n!} \sum_{k=1}^\infty \frac1{k^{a(n+1)}} \\ &= \frac1{\zeta(a)} \sum_{k=1}^\infty \frac1{k^a} \sum_{n=1}^\infty \frac{(-b)^n}{n!k^{an}} \\ &= \frac1{\zeta(a)} \sum_{k=1}^\infty \frac1{k^a} \big(e^{-b/k^a}-1\big). \end{align*} If $b$ is small, then this series is approximately $$ \frac1{\zeta(a)} \sum_{k=1}^\infty \frac1{k^a} \frac{-b}{k^a} = \frac{-b\zeta(2a)}{\zeta(a)}. $$ But if $b$ isn't that small, then the first few terms of the sum will be closer to $-1/k^a$, making the entire expression closer to $-1$.

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