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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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  • $\begingroup$ 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? $\endgroup$ – Stelios Sachpazis 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
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$$ \sum_{n\geq 1}\frac{H_{n-1}}{n}x^n = \frac{1}{2}\log(1-x)^2 \tag{1} $$ 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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  • $\begingroup$ 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. $\endgroup$ – Stelios Sachpazis 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? $\endgroup$ – Stelios Sachpazis 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
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\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(\frac12\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 $\zeta(2)=\frac{{\pi}^2}{6}$. $\endgroup$ – Stelios Sachpazis Apr 26 at 18:42

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