Is there a closed form for a give infinite sum?

Question

I've been asked to evaluate this sum $$\sum_{n=0}^{\infty}\frac{C_n^2}{2^{4n}}(H_{n+1}-H_n)$$ where $C_n=\frac1{n+1}{2n\choose n}$ denotes the $n$th Catalan number and $H_n$ denotes the nth Harmonic number. I'm wondering is this sum has a closed form.

Update

Using the idea of Jack D'Aurizio in this answer, I can write that

$$\sum_{n=0}^{\infty}\frac{C_n^2}{2^{4n}}(H_{n+1}-H_n)=\int_0^1 x K(x) \log^2(x)\mathrm{d}x$$

No as he stated in his solution $$ K(x)\stackrel{L^2(0,1)}{=}\sum_{n\geq 0}\frac{2}{2n+1}P_n(2x-1) $$

My question turn out to find the Fourier-Legendre expansion of the function $f(x)=x\log^2(x)$.

Answer 1

Your sum does have a closed form. On page 26 of New series involving harmonic numbers and squared central binomial coefficients by John Campbell your sum is given as

$$\sum_{n=0}^\infty\frac{\binom{2n}{n}^2}{16^n(n+1)^3}=\frac{48}{\pi}+16\ln(2)-\frac{32G}{\pi}-16,$$

where $G$ is Catalan's constant.

Unfortunately a full derivation is not given in the source.

Answer 2

A detailed proof. Let $S=\sum_{n=0}^{\infty}{\binom{2n}{n}^{2}\over 16^{n}(n+1)^3}$ be the sum we want to evaluate. If we consider the binomial expansions \begin{equation} 2-2\sqrt{1-x}=\sum_{n=0}^{\infty}\binom{2n}{n}{x^{n+1}\over 4^{n}(n+1)}\tag{1}, \end{equation} \begin{equation} -4\log{2}+4\log{\left(1+\sqrt{1-x}\right)}-4\sqrt{1-x}+4=\sum_{n=0}^{\infty}{\binom{2n}{n}x^{n+1}\over 4^n(n+1)^2}\tag{2}, \end{equation} where $(1)$ can be obtained from the binomial expansion of $(1-x)^{-1/2}$ after multiplying both sides by ${1 \over x}$, then integrating in $(0,x)$, and $(2)$ doing the same as with $(1)$. The change $x=k^{2}\sin^{2}x$ in $(2)$ gives \begin{equation} 4\log{\left({\sqrt{1-k^{2}\sin^{2}x}\over 2}+{1\over 2} \right)}-4\sqrt{1-k^{2}\sin^{2}x}+4=\sum_{n=0}^{\infty}{\binom{2n}{n}k^{2n+2}\sin^{2n+2}(x)\over 4^{n}(n+1)^{2}}\tag{3}, \end{equation} The next step we do is integrate $(3)$ in $(0,{\pi\over 2})$ and apply Wallis formula to obtain \begin{align} 2\pi\log\left({\sqrt{1-k^{2}}+1\over 4}\right)+4\int_{0}^{\pi \over 2}{x\cos{x}\over \sin{x}\sqrt{1-k^{2}\sin^{2}{x} }}dx-4E(k)+2\pi \\ \nonumber ={\pi\over 2}\sum_{n=0}^{\infty}{\binom{2n}{n}^{2}k^{2n+2}(n+1/2)\over 16^{n}(n+1)^{3}}\tag{4}, \end{align} where $E(k)$ is the complete elliptic integral of the second kind. Now we observe that $$S=2S_{1}-S_{2},$$ where $$S_{1}=\sum_{n=0}^{\infty}{\binom{2n}{n}^{2}\over 16^{n}(n+1)^{2}}\hspace{.5cm} and\hspace{.5cm} S_{2}=\sum_{n=0}^{\infty}{\binom{2n}{n}^{2}(2n+1)\over 16^{n}(n+1)^{3}}.$$ First we will see that $S_{1}={16\over \pi}-4$. See Calculate $\sum_{n = 0}^\infty \frac{C_n^2}{16^n}$ .

Finally setting $k=1$ in $(4)$ $$S_{2}=\sum_{n=0}^{\infty}{\binom{2n}{n}^{2}(2n+1)\over 16^{n}(n+1)^{3}}={4\over \pi}\left(-4\pi\log{2}+4\int_{0}^{\pi\over 2}{x\over \sin{x}}dx+2\pi-4\right) \\=-16\log{2}+{32G\over \pi}-{16\over \pi}+8,$$ where we have used the well known integral $\int_{0}^{\pi\over 2}{x\over \sin{x}}dx=2G$, being $G$ Catalan's constant. Then the desired sum is \begin{align*} S=2S_{1}-S_{2}=16\log{2}-{32G\over \pi}+{48\over \pi}-16. \end{align*} Bonus \begin{equation}_4F_3\left(a;b;x\right)=\sum_{n=0}^{\infty}{(a_{1})_{n}(a_{2})_{n}(a_{3})_{n}(a_{4})_{n}x^{n}\over (b_{1})_{n}(b_{2})_{n}(b_{3})_{n}n!} \end{equation} where $(a)_{n}$ is the Pochammer's symbol. If $x=1$: \begin{equation} \sum_{n=0}^{\infty}{\left({1\over 2}\right)^{2}_{n}(1)_{n}^{2}\over(2)_{n}^2(3)_{n} n!}=2\sum_{n=0}^{\infty}{\binom{2n}{n}^{2}\over 2^{4n}(n+2)(n+1)^{3}}=32\log{2}-{64G\over \pi}+{608\over 9\pi}-24, \end{equation} \begin{equation} \sum_{n=0}^{\infty}{\left({3\over 2}\right)^{2}_{n}(1)_{n}^{2}\over(2)_{n}(3)_{n}^2 n!}=16\sum_{n=1}^{\infty}{\binom{2n}{n}^{2}\over 2^{4n}n(n+1)^{2}}=64\log{2}-{128G\over \pi}-{320\over \pi}+96, \end{equation} where $G$ is Catalan's constant. \begin{equation} \sum_{n=0}^{\infty}{({1\over 4})_{n}^{2}({3\over 4})_{n}^{2}\over ({1\over 2})_{n}(1)_{n}({3\over 2})_{n}n!}=\sum_{n=0}^{\infty}{\binom{4n}{2n}^{2}\over 256^{n}(2n+1)}={2\over \pi}+{\sqrt{2\pi}\over 2\Gamma{(3/4)}^2}-{\sqrt{2}\Gamma{(3/4)}^2\over \pi^{3/2}}. \end{equation} \begin{equation} \sum_{n=0}^{\infty}{({1\over 4})_{n}^{2}({3\over 4})_{n}^{2}\over ({1\over 2})_{n}^2(2)_{n}n!}=\sum_{n=0}^{\infty}{\binom{4n}{2n}^{2}\over 256^{n}(n+1)}={20\over 9\pi}+{\sqrt{2\pi}\over 9\Gamma{(3/4)}^2}+{2\sqrt{2}\Gamma{(3/4)}^2\over 3\pi^{3/2}}, \end{equation} \begin{equation} \sum_{n=0}^{\infty}{({1\over 4})_{n}^{2}({3\over 4})_{n}^{2}\over(1)_{n} ({3\over 2})_{n}^2n!}=\sum_{n=0}^{\infty}{\binom{4n}{2n}^{2}\over 256^{n}(2n+1)^{2}}={8\over \pi}-{4\sqrt{2}\Gamma{(3/4)}^2\over \pi^{3/2}}, \end{equation} \begin{equation} \sum_{n=0}^{\infty}{(-{1\over 4})_{n}^{2}({1\over 4})_{n}^{2}\over({1\over 2})_{n} (1)_{n}^2n!}=\sum_{n=0}^{\infty}{\binom{4n}{2n}^{2}\over 256^{n}(4n-1)^{2}}={2\over \pi}+{\sqrt{2}\Gamma{(3/4)}^2\over \pi^{3/2}}, \end{equation} \begin{align} \sum_{n=0}^{\infty}{(-{3\over 4})_{n}^{2}(-{1\over 4})_{n}^{2}\over(-{1\over 2})_{n}^{2} (1)_{n}n!}={9\over 64}\sum_{n=0}^{\infty}{\binom{4n}{2n}^{2}\over 256^{n}(n+1)^{2}}+1={11\over 6\pi}+{\sqrt{2} \Gamma{(1/4)}^2\over12\pi^{3/2}}+{3\sqrt{2}\Gamma{(3/4)}^2\over 2\pi^{3/2}}-{3\sqrt{2} \Gamma{(1/4)}^2 \Gamma{(3/4)}^4\over 8 \pi^{7/2}}. \end{align} \begin{equation} \sum_{n=0}^{\infty}{\binom{4n}{2n}^{2}\over 256^{n}}\left({4\over (4n-1)^2}-{1\over (2n+1)^2}\right)={8\sqrt{2}\Gamma{(3/4)}^2\over \pi^{3/2}}, \end{equation}