Find the value of $A+B+C$ such that $$S=\sum_{n=1}^{n=\infty}\dfrac{1}{n^2\binom{2n}{n}}=\dfrac{A}{B}\zeta(C).$$

I solved it as follows: by using formula $$\sum_{n=1}^{n=\infty}\dfrac{1}{n\binom{2n}{n}}=\beta(n+1,n)=\beta(n,n+1)$$ $$S=\sum_{n=1}^{n=\infty}\dfrac{1}{n}\beta(n+1,n)$$ then i used definition of beta function to rewrite above sum as follows $$S=\sum_{n=1}^{n=\infty}\dfrac{1}{n}\int_{0}^{1} x^n(1-x)^{n-1}dx$$ then after changing order of integration and summation i again modified above as

$$S=\displaystyle\int_{0}^{1}\sum_{n=1}^{n=\infty}\left[\dfrac{x^n(1-x)^{n-1}}{n}\right]$$ from here i don't know how to proceed and relate it to zeta function (i don't no much about zeta function). Any help would be appreciated.

  • $\begingroup$ Welcom to MSE: Nice first question! $\endgroup$ – José Carlos Santos Jan 21 '18 at 9:32

More generally, the following power series expansion holds (see for example HERE): for $|x|\leq 1$ $$2(\arcsin(x))^2=\sum_{n=1}^{\infty} \frac{(2x)^{2n}}{n^2\binom{2n}{n}}.$$ Hence, for $x=1/2$, we find that $$\sum_{n=1}^{\infty} \frac{1}{n^2\binom{2n}{n}}=2(\arcsin(1/2))^2=2\left(\frac{\pi}{6}\right)^2=\frac{1}{3}\zeta(2).$$ Finally it easy to obtain $A+B+C=1+3+2=6$.

P.S. See also How to prove by arithmetical means that $\sum\limits_{k=1}^\infty \frac{((k-1)!)^2}{(2k)!} =\frac{1}{3}\sum\limits_{k=1}^{\infty}\frac{1}{k^{2}}$

  • $\begingroup$ @ Robert Z ....yeah putting x=1 on both sides will yield the answer....thank you S= pi^2/18 $\endgroup$ – Faraday Pathak Jan 21 '18 at 9:47
  • $\begingroup$ (A+B+C)=6.......i don't know radius of convergence $\endgroup$ – Faraday Pathak Jan 21 '18 at 9:49
  • $\begingroup$ See en.wikipedia.org/wiki/Radius_of_convergence In this case, it is the limit of $(n^2\binom{2n}{n})^{1/(2n)}$. In order to use the identity for $x=1$ you should check that $1<R$. $\endgroup$ – Robert Z Jan 21 '18 at 9:52
  • $\begingroup$ @ Robert Z...... it's radius of convergence(R) is coming 4 .....i calculated it using ratio test $\endgroup$ – Faraday Pathak Jan 21 '18 at 10:26
  • $\begingroup$ @veereshpandey Actually, the radius is $\sqrt{4}=2$, because $1/(n^2\binom{2n}{n})$ is the coefficient of $x^{2n}$. Anyway $R>1$. $\endgroup$ – Robert Z Jan 21 '18 at 10:55

The famous identity $$\zeta(2)=\sum_{n\geq 1}\frac{3}{n^2\binom{2n}{n}}$$ can be proved in many ways: creative telescoping, complex analysis, Lagrage's inversion theorem or Legendre polynomials, just to mention a few of them. Have a look at the first section of my notes.

A self-contained proof: $$\begin{eqnarray*}\sum_{n\geq 1}\frac{1}{n^2\binom{2n}{n}}=\sum_{n\geq 1}\frac{(n-1)!^2}{(2n)!}&=&\sum_{n\geq 1}\frac{\Gamma(n)^2}{2n\,\Gamma(2n)}\\&=&\sum_{n\geq 1}\frac{B(n,n)}{2n}\\&=&\sum_{n\geq 1}\frac{1}{2n}\int_{0}^{1}x^{n-1}(1-x)^{n-1}\,dx\\&=&-\frac{1}{2}\int_{0}^{1}\frac{\log(1-x+x^2)}{x(1-x)}\,dx\end{eqnarray*}$$ and now it is enough to notice that $\frac{1}{x(1-x)}=\frac{1}{x}+\frac{1}{1-x}$ and that $1-x+x^2$ is a cyclotomic polynomial, in order to exploit this lemma: $$ \int_{0}^{1}\frac{\log\Phi_n(x)}{x}\,dx = \frac{\zeta(2)(-1)^{\omega(n)+1}\varphi(n)\,\text{rad}(n)}{n^2}.$$

  • $\begingroup$ @ Jack D'Aurizio...........your approach is great sir....thank you so much.....but i have to say that i am unable to understand "the lemma(which you exploited)" above because i'm not well acquainted with Mobius inversion .... $\endgroup$ – Faraday Pathak Jan 21 '18 at 18:08
  • 1
    $\begingroup$ @veereshpandey: actually you do not need Moebius inversion to tackle this specific problem. Just $1-x+x^2 = \frac{1+x^3}{1+x}$ and the fact that $\int_{0}^{1}\frac{\log(1+x^a)}{x}\,dx$ is related to $\zeta(2)$ in a simple way. Substitute $x=z^{1/a}$ and you are done. $\endgroup$ – Jack D'Aurizio Jan 21 '18 at 18:18
  • $\begingroup$ @ Jack D'Aurizio ......after doing partial fractions(as you've pointed out in answer) and manipulating argument of 'log' as you said (i'm able to solve 2 integrals having x in denominator in terms of zeta(2))....but, i'm still left with two integrals having "1-x" in denominator (how to tackle them?) with limits 0 to 1. $\endgroup$ – Faraday Pathak Jan 21 '18 at 18:56
  • 1
    $\begingroup$ @veereshpandey: $1-x+x^2$ is left unchanged by the substitution $x\mapsto 1-x$. $\endgroup$ – Jack D'Aurizio Jan 21 '18 at 18:57
  • $\begingroup$ @ Jack D'Aurizio .....oh ,yes..,,thank you sir. $\endgroup$ – Faraday Pathak Jan 21 '18 at 19:02

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

I'll start from de Jack D'Aurizio "integral expression":

\begin{align} S & = -\,{1 \over 2}\int_{0}^{1}{\ln\pars{1 - x + x^{2}} \over x\pars{1 - x}}\,\dd x = -\,{1 \over 2}\int_{0}^{1} {\ln\pars{1 - x + x^{2}} \over x}\,\dd x - {1 \over 2}\ \overbrace{\int_{0}^{1}{\ln\pars{1 - x + x^{2}} \over 1 - x}\,\dd x} ^{\ds{\mbox{lets}\ x\ \mapsto\ 1 - x}} \\[5mm] & = -\int_{0}^{1}{\ln\pars{1 - x + x^{2}} \over x}\,\dd x \,\,\,\stackrel{\mrm{IBP}}{=}\,\,\, \int_{0}^{1}\ln\pars{x}\,{2x - 1 \over x^{2} - x + 1}\,\dd x = \int_{0}^{1}\ln\pars{x}\,{2x - 1 \over \pars{x - r}\pars{x - \bar{r}}}\,\dd x \end{align}

where $\ds{r \equiv {1 + \root{3}\ic \over 2} = \exp\pars{{\pi \over 3}\,\ic}}$.

Then, \begin{align} S & = \int_{0}^{1}\ln\pars{x}\bracks{{2r - 1 \over \pars{r - \bar{r}}\pars{x - r}} -{2\bar{r} - 1 \over \pars{r - \bar{r}}\pars{x - \bar{r}}}}\dd x = {1 \over 2\ic\,\Im\pars{r}}\bracks{2\ic\,\Im\int_{0}^{1}\ln\pars{x} \,{2r - 1 \over x - r}\,\dd x} \\[5mm] & = {1 \over \root{3}/2}\,\Im\pars{\bracks{-\root{3}\ic} \int_{0}^{1}{\ln\pars{x} \over r - x}\,\dd x} = -2\,\Re \int_{0}^{1/r}{\ln\pars{rx} \over 1 - x}\,\dd x \\[5mm] & \stackrel{\mrm{IBP}}{=}\,\,\, -2\,\Re\int_{0}^{\large\bar{r}}{\ln\pars{1 - x} \over x}\,\dd x = 2\,\Re\int_{0}^{\large\bar{r}}\mrm{Li}_{2}'\pars{x}\,\dd x = 2\,\Re\mrm{Li}_{2}\pars{\exp\pars{-\,{\pi \over 3}\,\ic}} \\[5mm] & = 2\,\Re\mrm{Li}_{2}\pars{\exp\pars{2\pi\bracks{1 \over 6}\,\ic}} \\[5mm] & = -\,{\pars{2\pi\ic}^{2} \over 2!}\,\ \overbrace{\mrm{B}_{2}\pars{1 \over 6}}^{\ds{1 \over 36}} = {\pi^{2} \over 18}\qquad\qquad \pars{~\mrm{B}_{n}\pars{x}:\ Bernoulli\ Polynomial~} \end{align}

which is Junqui$\mathrm{\grave{e}}$re's Inversion Formula in terms of Bernoulli Polynomials $\ds{\mrm{B}_{n}\pars{x}}$. Note that $\ds{\mrm{B}_{2}\pars{x} = x^{2} - x + {1 \over 6}}$.

Then, $$ \bbx{S \equiv \sum_{n = 1}^{\infty}{1 \over n^{2}{2n \choose n}} = \zeta\pars{2}\,{1 \over 3}} $$


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.