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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 $f(x)=x\log^2(x)$.

Blue
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Arnaud
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2 Answers2

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As @Angina Seng has observed: $H_{n+1}-H_{n}=\frac{1}{n+1}$ and then your sum is $S=\sum_{n=0}^{\infty}{\binom{2n}{n}^{2}\over 16^{n}(n+1)^3}$ 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*}

User-Refolio
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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.

dxdydz
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    Alternatively, $$\sum_{n=0}^\infty\frac{\binom{2n}{n}^2}{16^n(n+1)^3}= ,4F_3,\big( {\tfrac12,\tfrac12,1,1};,{2,2,2};,1\big)$$ This binomial sum reminds me a bit of the logsine integral, $$\sum{n=0}^\infty\frac{\binom{2n}{n}}{16^n(2n+1)^3}= ,_4F_3,\big( {\tfrac12,\tfrac12,\tfrac12,\tfrac12};,{\tfrac32,\tfrac32,\tfrac32};,\tfrac14\big)\=\tfrac12\int_0^{\pi/3}\ln^2(2\sin \tfrac{x}2\big)dx = \tfrac{7}{216}\pi^3$$ – Tito Piezas III Jun 07 '19 at 02:57
  • @TitoPiezasIII Below a proof. Best all! – User-Refolio Nov 01 '23 at 12:34