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


To proceed I need to evaluate the


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.

  • $\begingroup$ The last integral looks like the incomplete beta function (for $k >0$). $\endgroup$ – Claude Leibovici Jun 23 '17 at 14:54
  • $\begingroup$ @ClaudeLeibovici Yes, it does. Can we find values of the incomplete beta function? $\endgroup$ – Στέλιος Σαχπάζης Jun 23 '17 at 14:56
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    $\begingroup$ As far as I know (and I know little, be sure), there is no explicit formula for $B_{\frac{1}{2}}(2 k,0)$ $\endgroup$ – Claude Leibovici Jun 23 '17 at 15:01

$$ \sum_{n\geq 1}\frac{H_{n-1}}{n}x^n = \frac{1}{2}\log(1-x)^2 \tag{1} $$ 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.

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.

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  • $\begingroup$ Yes, but the dilogarithm reflection formula, in your proof, uses the fact that $\displaystyle{\zeta (2)=\frac{{\pi}^2}{6}}.$ Am I wright? I ended up with this series when I was trying to compute $\zeta(2)$ with a certain method. $\endgroup$ – Στέλιος Σαχπάζης Jun 23 '17 at 15:14
  • $\begingroup$ @SachpazisStelios: correct. Have a look at this thread (math.stackexchange.com/questions/8337/…) for proving $\zeta(2)=\frac{\pi^2}{6}$. $\endgroup$ – Jack D'Aurizio Jun 23 '17 at 15:19
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    $\begingroup$ Is $\text{Li}_2$ really 'real analytic'? Invoking the polylogarithm feels like cheating the 'elementary way' a bit, indirectly invoking Riemann's zeta. $\endgroup$ – orlp Jun 23 '17 at 15:21
  • $\begingroup$ @Jack D'Aurizio I knew this amazing thread. Do you remember if there is anyone in this thread, trying to evaluate $\zeta(2)$, like I did? $\endgroup$ – Στέλιος Σαχπάζης Jun 23 '17 at 15:22
  • $\begingroup$ @SachpazisStelios: not through the series you want to use, but I showed (and Euler well before me) that by accelerating $\sum_{n\geq 1}\frac{1}{n^2}$ as $\sum_{n\geq 1}\frac{3}{n^2\binom{2n}{n}}$ by creative telescoping, the value of $\zeta(2)$ depends on $\arcsin\left(\frac{1}{2}\right)^2$. $\endgroup$ – Jack D'Aurizio Jun 23 '17 at 15:30

\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}

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  • $\begingroup$ If you read carefully my question, I proved this. Now, why $\zeta(2)=\frac{{\pi}^2}{6}$ by elementary methods? I want to calculate the series in an other way, to prove that $\zeta(2)=\frac{{\pi}^2}{6}$. $\endgroup$ – Στέλιος Σαχπάζης Apr 26 '19 at 18:42
  • $\begingroup$ @Stelios Sachpazis sorry didnt pay attention to that. I submitted a proof to your inquiry. $\endgroup$ – Ali Shadhar Oct 24 '19 at 5:35
  • $\begingroup$ +1. At the end of the first line: $\displaystyle\left(1 \over 2\right)^{n}$ must be $\displaystyle\left(\color{red}{\large x} \over 2\right)^{n}$. Otherwise, everything is fine. $\endgroup$ – Felix Marin Jul 27 at 2:46
  • $\begingroup$ Thank you @Felix Martin $\endgroup$ – Ali Shadhar Jul 27 at 5:54

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


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  • $\begingroup$ This also provides a second way of calculating the integral $\int_{0}^{\pi/2}\ln (2\cos x)dx.$ $\endgroup$ – Στέλιος Σαχπάζης Feb 24 at 14:08

\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:

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

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