# Find the value of $A+B+C$ in the following question?

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.

• Welcom to MSE: Nice first question! – 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 $A+B+C=1+3+2=6$.

• @ Robert Z ....yeah putting x=1 on both sides will yield the answer....thank you S= pi^2/18 – Faraday Pathak Jan 21 '18 at 9:47
• (A+B+C)=6.......i don't know radius of convergence – Faraday Pathak Jan 21 '18 at 9:49
• 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 $1<R$. – Robert Z Jan 21 '18 at 9:52
• @ Robert Z...... it's radius of convergence(R) is coming 4 .....i calculated it using ratio test – Faraday Pathak Jan 21 '18 at 10:26
• @veereshpandey Actually, the radius is $\sqrt{4}=2$, because $1/(n^2\binom{2n}{n})$ is the coefficient of $x^{2n}$. Anyway $R>1$. – 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: $$\int_{0}^{1}\frac{\log\Phi_n(x)}{x}\,dx = \frac{\zeta(2)(-1)^{\omega(n)+1}\varphi(n)\,\text{rad}(n)}{n^2}.$$

• @ 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 .... – Faraday Pathak Jan 21 '18 at 18:08
• @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. – Jack D'Aurizio Jan 21 '18 at 18:18
• @ 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. – Faraday Pathak Jan 21 '18 at 18:56
• @veereshpandey: $1-x+x^2$ is left unchanged by the substitution $x\mapsto 1-x$. – Jack D'Aurizio Jan 21 '18 at 18:57
• @ Jack D'Aurizio .....oh ,yes..,,thank you sir. – Faraday Pathak Jan 21 '18 at 19:02


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