Elementary way to calculate the series $\sum\limits_{n=1}^{\infty}\frac{H_n}{n2^n}$

Question

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.

Answer 1

$$ \sum_{n\geq 1}\frac{H_{n-1}}{n}x^n = \frac{1}{2}\log(1-x)^2 \tag{1} $$ follows from the termwise integration of $\sum_{n\geq 1}H_n x^n = \frac{-\log(1-x)}{1-x}.$ It leads to $$ \sum_{n\geq 1}\frac{H_{n-1}}{n 2^n} = \frac{1}{2}\log^2(2).\tag{2}$$ On the other hand $$ \sum_{n\geq 1}\frac{1}{n^2 2^n} = \text{Li}_2\left(\frac{1}{2}\right)=\frac{\pi^2}{12}-\frac{\log^2(2)}{2}\tag{3} $$ follows from the dilogarithm reflection formula, and by summing $(2)$ and $(3)$ the identity we usually derive from Euler's acceleration method $$ \sum_{n\geq 1}\frac{H_n}{n 2^n} = \frac{\pi^2}{12}\tag{4} $$ easily follows.

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Late addendum (2020). It is interesting to notice that $$ H_n = \gamma-\lim_{m\to 0^+} \frac{d}{dm}\left(\frac{1}{(n+1)_m}\right) \tag{5}$$ where $(n+1)_m$ is the rising Pochhammer symbol and $\gamma=-\Gamma'(1)$ is the Euler-Mascheroni constant.
We have $$\begin{eqnarray*} \sum_{n\geq 1}\frac{1}{n 2^n (n+1)_m}&=&\sum_{n\geq 1}\frac{1}{2^n(n)_{m+1}}=\sum_{n\geq 1}\frac{\Gamma(n)}{2^n \Gamma(n+m+1)}\\&=&\frac{1}{\Gamma(m+1)}\sum_{n\geq 1}\frac{B(n,m+1)}{2^n}\\&=&\frac{1}{\Gamma(m+1)}\int_{0}^{1}\sum_{n\geq 1}\frac{(1-x)^m x^{n-1}}{2^n}\,dx\\&=&\frac{1}{\Gamma(m+1)}\int_{0}^{1}\frac{(1-x)^m}{2-x}\,dx\\&=&\frac{1}{\Gamma(m+1)}\int_{0}^{1}\frac{x^m}{1+x}\,dx\end{eqnarray*} \tag{6}$$ hence by the chain rule and differentiation under the integral sign $$\begin{eqnarray*} \sum_{n\geq 1}\frac{H_n}{n 2^n}&=&\gamma\log(2)-\gamma\int_{0}^{1}\frac{dx}{1+x}-\int_{0}^{1}\frac{\log(x)}{1+x}\,dx\\&=&\int_{0}^{1}-\log(x)\sum_{n\geq 0}(-1)^{n} x^n\,dx=\sum_{n\geq 0}\frac{(-1)^n}{(n+1)^2}\\&=&\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^2}=\zeta(2)-2\sum_{n\geq 1}\frac{1}{(2n)^2}=\left(1-\frac{1}{2}\right)\zeta(2)=\color{red}{\frac{\pi^2}{12}}\end{eqnarray*} \tag{7}$$ "undoing" Euler acceleration method.

Answer 2

\begin{align} \sum_{n=1}^{\infty}\frac{H_n}{2^n n}&=\sum_{n=1}^{\infty}\frac{H_n}{2^n}\int_o^1 x^{n-1}\ dx=\int_0^1\frac{1}{x}\sum_{n=1}^{\infty}\left(\frac x2\right)^nH_n\ dx\\ &=-\int_{0}^{1}\frac{\ln(1-x/2)}{x(1-x/2)}\ dx\overset{x\mapsto2x}{=}-\int_{0}^{1/2}\frac{\ln(1-x)}{x(1-x)}\ dx\\ &=-\int_{0}^{1/2}\frac{\ln(1-x)}{x}\ dx-\int_{0}^{1/2}\frac{\ln(1-x)}{1-x}\ dx\\ &= \operatorname{Li_2}(1/2)+\frac12\ln^22=\frac12\zeta(2)-\frac12\ln^22+\frac12\ln^22=\frac12\zeta(2) \end{align}

Answer 3

I found the proof that $\zeta(2)=\frac{\pi^2}{6}$ on YouTube but I did little changes:

\begin{align} I&=\int_0^{\pi/2}\ln(2\cos x)\ dx=\int_0^{\pi/2}\ln\left(e^{ix}(1+e^{-2ix})\right)\ dx\\ &=\int_0^{\pi/2}ix\ dx-\sum_{n=1}^\infty \frac{(-1)^n}{n}\int_0^{\pi/2}e^{-2ix}\ dx\\ &=\frac{\pi^2}{8}i-\sum_{n=1}^\infty\frac{(-1)^n}{n}\left(-\frac{(-1)^n-1}{2in}\right)\\ &=\frac{\pi^2}{8}i-\frac12i\left(\zeta(2)-\operatorname{Li}_2(-1)\right)\\ &=\frac{\pi^2}{8}i-\frac12i\left(\zeta(2)+\frac12\zeta(2)\right)\\ &=i\left(\frac{\pi^2}{8}-\frac34\zeta(2)\right) \end{align}

By comparing the imaginary parts, we have

$$0=\frac{\pi^2}{8}-\frac34\zeta(2)\Longrightarrow\zeta(2)=\frac{\pi^2}{6}$$

Answer 4

\begin{align}J&=\int_{0}^{\frac{1}{2}}\frac{\ln (1-x)}{x(x-1)}dx\\ &\overset{x=\frac{y}{1+y}}=\int_0^1 \frac{\ln(1+y)}{y}\,dy\\ &=\int_0^1 \frac{\ln(1-y^2)}{y}\,dy-\int_0^1 \frac{\ln(1-t)}{t}\,dt\\ &\overset{u=y^2}=\frac{1}{2}\int_0^1 \frac{\ln(1-u)}{u}\,du-\int_0^1 \frac{\ln(1-t)}{t}\,dt\\ &=-\frac{1}{2}\int_0^1 \frac{\ln(1-u)}{u}\,du\\ &\overset{w=1-u}=-\frac{1}{2}\int_0^1 \frac{\ln w}{1-w}\,dw\\ &=-\frac{1}{2}\times -\frac{\pi^2}{6}\\ &=\boxed{\frac{\pi^2}{12}} \end{align}

NB: i assume that:

$\displaystyle \int_0^1 \frac{\ln x}{1-x}\,dx=-\zeta(2)=-\frac{\pi^2}{6}$