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

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.

Answer 1

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 $\sum\limits_{k=1}^\infty \frac{((k-1)!)^2}{(2k)!} =\frac{1}{3}\sum\limits_{k=1}^{\infty}\frac{1}{k^{2}}$

Answer 2

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: $$ \int_{0}^{1}\frac{\log\Phi_n(x)}{x}\,dx = \frac{\zeta(2)(-1)^{\omega(n)+1}\varphi(n)\,\text{rad}(n)}{n^2}.$$

Answer 3

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