7
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ The last integral looks like the incomplete beta function (for $k >0$). $\endgroup$ Jun 23, 2017 at 14:54
  • $\begingroup$ @ClaudeLeibovici Yes, it does. Can we find values of the incomplete beta function? $\endgroup$ Jun 23, 2017 at 14:56
  • 1
    $\begingroup$ As far as I know (and I know little, be sure), there is no explicit formula for $B_{\frac{1}{2}}(2 k,0)$ $\endgroup$ Jun 23, 2017 at 15:01

4 Answers 4

5
$\begingroup$

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

$\endgroup$
6
  • $\begingroup$ 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. $\endgroup$ Jun 23, 2017 at 15:14
  • $\begingroup$ @SachpazisStelios: correct. Have a look at this thread (math.stackexchange.com/questions/8337/…) for proving $\zeta(2)=\frac{\pi^2}{6}$. $\endgroup$ Jun 23, 2017 at 15:19
  • 1
    $\begingroup$ 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, 2017 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? $\endgroup$ Jun 23, 2017 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 $\arcsin\left(\frac{1}{2}\right)^2$. $\endgroup$ Jun 23, 2017 at 15:30
2
$\begingroup$

\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}

$\endgroup$
4
  • $\begingroup$ 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 $\zeta(2)=\frac{{\pi}^2}{6}$. $\endgroup$ Apr 26, 2019 at 18:42
  • $\begingroup$ @Stelios Sachpazis sorry didnt pay attention to that. I submitted a proof to your inquiry. $\endgroup$ Oct 24, 2019 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. $\endgroup$ Jul 27, 2020 at 2:46
  • $\begingroup$ Thank you @Felix Martin $\endgroup$ Jul 27, 2020 at 5:54
1
$\begingroup$

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

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

$\endgroup$
1
  • $\begingroup$ This also provides a second way of calculating the integral $\int_{0}^{\pi/2}\ln (2\cos x)dx.$ $\endgroup$ Feb 24, 2020 at 14:08
0
$\begingroup$

\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:

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.