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Manipulating the gamma function, I realized that this result is equal to $\eta^{\prime}(1)$, where $\eta(x)$ is the Dirichet's eta function. $\gamma$ is the Euler-Mascheroni constant.

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    $\begingroup$ Here are found some proofs: artofproblemsolving.com/community/q1h1761372 $\endgroup$ – Zacky Jan 23 at 21:14
  • $\begingroup$ The first step is $\zeta(s)-\frac1{s-1} = (s-1) \int_1^\infty (\sum_{n \le x}\frac1n - \log x) x^{-s}dx=(s-1) \int_1^\infty(\gamma+O(x^{-1}))x^{-s}dx= \gamma+O(\frac{s-1}{s})$. The next step is to find $\eta'(1)$ in term of $f'(1)=\gamma$ where $\eta(s) = (1-2^{1-s}) \zeta(s), f(s) =(s-1)\zeta(s)$ $\endgroup$ – reuns Jan 23 at 21:34

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