Prove that $\sum\limits_{n=0}^\infty\binom{2n}n\frac n{4^n(n+1)^2}=\ln(16)-2$

Question

Prove that $$\sum\limits_{n=0}^\infty\binom{2n}n\frac n{4^n(n+1)^2}=\ln(16)-2$$

According to Wolfram the above holds. Could someone show me the steps for this?

Answer 1

HINT

First note that $$\sum_{n\geq0}\binom{2n}n\frac n{4^n(n+1)^2}=\sum_{n\geq0}\binom{2n}n\frac{(n+1)-1}{4^n(n+1)^2}=\sum_{n\geq0}\binom{2n}n\frac1{4^n(n+1)}-\sum_{n\geq0}\binom{2n}n\frac1{4^n(n+1)^2}$$ Now recall that $$\sum_{n\geq0}\binom{2n}n\frac{x^n}{4^n}=\frac1{\sqrt{1-x}},~~~\text{for }|x|<1$$ The task boils down to an exercise in integration. Can you take it from here?

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For the interested reader I will add the complete solution now. As given, note that for the first sum we have the following $$\sum_{n\geq0}\binom{2n}n\frac1{4^n(n+1)}=\sum_{n\geq0}\binom{2n}n\frac1{4^n}\int_0^1x^n{\rm d}x=\int_0^1\frac{{\rm d}x}{\sqrt{1-x}}=[-2\sqrt{1-x}]_0^1=2$$ This also gives us the following identity $$\sum_{n\geq0}\binom{2n}n\frac{x^n}{4^n(n+1)}=\int\frac{{\rm d}x}{\sqrt{1-x}}=-2\sqrt{1-x}+c$$ Take $x=0$ to see that $c=2$. Dividing by $x$ and integrating again over $[0;1]$ gives \begin{align*} \sum_{n\geq0}\binom{2n}n\frac1{4^n(n+1)^2}&=\int_0^1\frac{2-2\sqrt{1-x}}x{\rm d}x\\ &=2\int_0^1\frac{1-\sqrt{1-x}}x{\rm d}x\\ &=2\int_0^1\frac{1-\sqrt x}{1-x}{\rm d}x\\ &=2\int_0^1\frac1{1+\sqrt x}{\rm d}x\\ &=2\sum_{n\geq0}(-1)^n\int_0^1x^{\frac n2}{\rm d}x\\ &=2\sum_{n\geq0}\frac{(-1)^n}{\frac n2+1}\\ &=4\sum_{n\geq0}\frac{(-1)^n}{n+2}\\ &=4\left[\sum_{n\geq1}\frac{(-1)^n}n+1\right]\\ &=4-4\log2 \end{align*} The result follows.

$$\therefore~\sum_{n\geq0}\binom{2n}n\frac n{4^n(n+1)^2}~=~4\log2-2$$

Answer 2

The $n$-th Catalan number is defined as $$ C_n := \frac 1 {n+1}\binom{2n}n.$$ By inspecting the family of definite integrals $$ J_n(\alpha) := \frac 1 \pi \int_0^\alpha \xi^{2n} \sqrt{\alpha^2-\xi^2} d\xi, \qquad \alpha > 0,\ n\geq 0, $$ it can be shown, via standard integration techniques, that $C_n = J_n(2)$: equivalently, through the substitution $x=\xi^2$, $$ C_n = \frac{1}{2\pi} \int_0^4 x^n \sqrt{\frac{4-x}{x}} dx.$$ Then your sum $S$ becomes $$\begin{split} S &=\sum_{n=0}^\infty \frac{ n C_n}{(n+1) 4^n } = \sum_{n=0}^\infty \frac{(n+1-1) C_n}{(n+1)4^n} = \sum_{n=0}^\infty \frac{C_n}{4^n} - \sum_{n=0}^\infty \frac{C_n}{4^n(n+1)} \\ &= \frac{1}{2\pi} \int_0^4 \sqrt{\frac{4-x}{x}} \left[\sum_{n=0}^\infty \left( \frac{x}{4}\right)^n -\sum_{n=0}^\infty \frac{(x/4)^n}{n+1} \right] dx \\ &= \frac{1}{2\pi} \int_0^4 \sqrt{\frac{4-x}{x}} \left[ \frac{1}{1-x/4} + \frac{\log(1-x/4)}{x/4} \right]dx \\ &= \frac{1}{2\pi} \int_0^4 \sqrt{\frac{4-x}{x}} \left[ \frac{4}{4-x}+\frac 4 x \log\left(\frac{4-x}{4} \right) \right] dx \\ &=\frac{2}{\pi} \int_0^4 \frac{dx}{\sqrt{4-(x-2)^2}} + \frac 2 \pi \int_0^4 \frac 1 x \log\left(\frac{4-x}{4} \right) dx \\ &= \frac 2\pi \int_0^\pi dt + \frac{2}{\pi} \int_0^1 \frac{\log s}{1-s} \sqrt{\frac{s}{1-s}}ds \\ &=: 2+\frac 2\pi I, \end{split}$$ where we have interchanged sum and integral by total convergence of power series, and employed the formula for geometric series and its integral, and parallel substitutions $x = 2-2\cos t$, $s = 1 - x/4$. The definite integral $I$ can be resolved as follows: $$\begin{split} I &= \int_0^1 \frac{\log s}{1-s} \sqrt{\frac{s}{1-s}}ds=\int_0^\infty \log \left( \frac y {y+1}\right) \frac{\sqrt y}{y+1} dy = 2\int_0^\infty \log \left( \frac {x^2} {x^2+1}\right) \frac{x^2}{x^2+1} dx \\ &= \underbrace{\left[(x-\arctan x)\log \left( \frac {x^2} {x^2+1}\right) \right]_{-\infty}^{+\infty}}_{=0} -2 \int_{-\infty}^\infty \frac{x-\arctan x}{x(1+x^2)}dx \\ &= -2\pi +2 \int_{-\pi/2}^{\pi/2} z \cot(z) dz = -2\pi + 4 \left(\underbrace{[z \log(\sin z)]_0^{\pi/2}}_{=0} - \int_0^{\pi/2} \log(\sin z) dz \right), \end{split}$$ where we have employed substitutions $y = s/(1-s)$, $x = \sqrt y$, and then $z = \arctan x$. The remaining integral has been evaluated many times over on MSE (see this answer): in total, we have $$I = 2\pi(\log 2 - 1), \quad \implies \quad S = 2 + 4 (\log 2 -1) = \boxed{\log(16)-2} $$ as required.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{% \sum_{n = 0}^{\infty}{2n \choose n}{n \over 4^{n}\pars{n + 1}^{2}}} = \sum_{n = 0}^{\infty}\bracks{{-1/2 \choose n}\pars{-4}^{n}}{1 \over 4^{n}} \bracks{{1 \over n + 1} - {1 \over \pars{n + 1}^{2}}} \\[5mm] = & \left.\pars{1 + \partiald{}{a}}\sum_{n = 0}^{\infty}{-1/2 \choose n} {\pars{-1}^{n} \over n + a + 1}\,\right\vert_{\large\ a\ =\ 0} \\[5mm] = & \left.\pars{1 + \partiald{}{a}}\sum_{n = 0}^{\infty}{-1/2 \choose n} \pars{-1}^{n}\int_{0}^{1}t^{n + a}\,\dd t\,\right\vert_{\large\ a\ =\ 0} \\[5mm] = & \left.\pars{1 + \partiald{}{a}}\int_{0}^{1}t^{a} \sum_{n = 0}^{\infty}{-1/2 \choose n}\pars{-t}^{n}\,\dd t\,\right\vert_{\large\ a\ =\ 0} \\[5mm] = &\ \left.\pars{1 + \partiald{}{a}}\int_{0}^{1}t^{a}\pars{1 - t}^{-1/2} \,\dd t\,\right\vert_{\large\ a\ =\ 0} \\[5mm] = &\ \left.\pars{1 + \partiald{}{a}}{\Gamma\pars{1 + a}\Gamma\pars{1/2} \over \Gamma\pars{3/2 + a}}\,\right\vert_{\large\ a\ =\ 0} \\[5mm] = &\ \underbrace{{\Gamma\pars{1}\Gamma\pars{1/2} \over \Gamma\pars{3/2}}} _{\ds{2}}\ +\ \underbrace{\left.\partiald{}{a}{\Gamma\pars{1 + a}\Gamma\pars{1/2} \over \Gamma\pars{3/2 + a}}\,\right\vert_{\large\ a\ =\ 0}} _{\ds{-4 + \ln\pars{16}}} \\[5mm] = &\ \bbx{\ln\pars{16} - 2} \approx 0.7726 \\ & \end{align}

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$\underline{Note\ that}$ \begin{align} &\bbox[5px,#ffd]{\left.\partiald{}{a}{\Gamma\pars{1 + a}\Gamma\pars{1/2} \over \Gamma\pars{3/2 + a}} \,\right\vert_{\ a\ =\ 0}} = \bracks{a^{1}}{\Gamma\pars{1 + a}\Gamma\pars{1/2} \over \Gamma\pars{3/2 + a}} \\[5mm] = &\ \Gamma\pars{1 \over 2}\bracks{a^{1}} {\Gamma\pars{1} + \Gamma'\pars{1}a \over \Gamma\pars{3/2} + \Gamma'\pars{3/2}a} = \Gamma\pars{1 \over 2}{1 \over \Gamma\pars{3/2}}\bracks{a^{1}} {1 + \Psi\pars{1}a \over 1 + \Psi\pars{3/2}a} \\[5mm] = &\ 2\bracks{a^{1}} \pars{1 - \gamma a}\bracks{1 - \Psi\pars{3 \over 2}a} = -2\bracks{\gamma + \Psi\pars{1 \over 2} + 2} \\[5mm] = &\ -4 - 2\int_{0}^{1}{1 - t^{-1/2} \over 1 - t}\,\dd t = -4 - 2\int_{0}^{1}{1 - t^{-1} \over 1 - t^{2}}\,2t\,\dd t = -4 + 4\int_{0}^{1}{\dd t \over 1 + t} \\[5mm] = &\ -4 + 4\ln\pars{2} = \bbx{-4 + \ln\pars{16}} \\ & \end{align}