# How to show $\sum_{n=2}^{\infty}(-1)^n \frac{\log(n)}{n} = \gamma \log(2) - \frac{\log^2(2)}{2}$?

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.

• Here are found some proofs: artofproblemsolving.com/community/q1h1761372 – Zacky Jan 23 at 21:14
• 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 $\eta(s) = (1-2^{1-s}) \zeta(s), f(s) =(s-1)\zeta(s)$ – reuns Jan 23 at 21:34