I want to calculate the series of the Basel problem $\displaystyle{\sum_{n=1}^{\infty}\frac{1}{n^2}}$ by applying the Euler series transformation. With some effort I got that
$$\displaystyle{\frac{\zeta (2)}{2}=\sum_{n=1}^{\infty}\frac{H_n}{n2^n}}.$$
I know that series like the $\displaystyle{\sum_{n=1}^{\infty}\frac{H_n}{n2^n}}$ are evaluated here, but the evaluations end up with some values of the $\zeta$ function, like $\zeta (2),\zeta(3).$
First approach: Using the generating function of the harmonic numbers and integrating term by term, I concluded that
$$\displaystyle{\sum_{n=1}^{\infty}\frac{H_n}{n2^n}=\int_{0}^{\frac{1}{2}}\frac{\ln (1-x)}{x(x-1)}dx},$$
but I can't evaluate this integral with any real-analytic way.
First question: Do you have any hints or ideas to evaluate it with real-analytic methods?
Second approach: I used the fact that $\displaystyle{\frac{H_n}{n}=\sum_{k=1}^{n}\frac{1}{k(n+k)}}$ and then, I changed the order of summation to obtain
$$\displaystyle{\sum_{n=1}^{\infty}\frac{H_n}{n2^n}=\sum_{k=1}^{\infty}\frac{2^k}{k}\left(\sum_{m=2k}^{\infty}\frac{1}{m2^m}\right)}.$$
To proceed I need to evaluate the
$$\int_{0}^{\frac{1}{2}}\frac{x^{2k-1}}{1-x}dx,$$
since $\displaystyle{\sum_{m=2k}^{\infty}\frac{1}{m2^m}=\int_{0}^{\frac{1}{2}}\frac{x^{2k-1}}{1-x}dx}.$
Second question: How can I calculate this integral?
Thanks in advance for your help.
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