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I would like to get an expression which represents the series $$\sum_{n\geq1}(-1)^{n+1}\frac{\text{H}_{n}\zeta(1+n)}{n+1}$$ in terms of $\zeta (2)$ and a series expansion containing $\zeta (n)$. Here, $\text{H}_{n}=\sum_{k=1}^n 1/k $ stands for the harmonic number, and $\zeta$ is the Riemann zeta function. I am aware of the equalities: $$ \sum _{n=2}^{\infty} H_n( \zeta(n)-\zeta(n+1))={\pi}^2/6-\gamma$$ and $$\gamma=\sum_{n=2}^{\infty} \frac {(-1)^n\zeta (n)}{n},$$ where $\gamma$ is the Euler constant. But I do not sure if these could be related to the series under the question.

Thanks to any one who would be kind as to give a help.

Masacroso
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karparvar
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    You can rewrite the series that is present in the text of the question as $$\frac{1}{2}\sum_{m\geq1}\log^{2}\left(1+\frac{1}{m}\right).$$ Now just take a look at the following question to get $\zeta\left(2\right)-\sum_{n\geq1}\frac{\zeta\left(2n\right)}{2n^{2}}$ https://math.stackexchange.com/questions/2311735/a-closed-form-of-sum-n-1-infty-1n-ln-left1-frac12n-right?noredirect=1&lq=1 – Marco Cantarini Jul 27 '21 at 11:59
  • Using the representation $H_ {n}=\int_{0}^{1}\frac{1-x^n}{1-x},dx$ the sum can also be written as $\int_{0}^{1}\frac{ \log(\Gamma(1+x))}{x(-1+x)},dx$ – Dr. Wolfgang Hintze Jul 28 '21 at 20:33

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