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By Mathematica, we find $$\sum_{n=1}^\infty \frac{4^n}{n^3\binom{2n}{n}}=\pi^2\log(2)-\frac{7}{2}\zeta(3).$$

How to find the closed form for general series: $$\sum_{n=1}^\infty \frac{4^n}{n^p\binom{2n}{n}}? \ \ (p\ge 3)$$

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    $\begingroup$ What do you mean by "By Mathematics"? $\endgroup$ – Adam Chalumeau Jun 3 '19 at 10:21
  • $\begingroup$ I guess that, if $p>3$, the result is an hypergeometric function and probably no closed form (but I may be wrong). $\endgroup$ – Claude Leibovici Jun 3 '19 at 10:21
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    $\begingroup$ Adam Chalumeau ``By Mathematica'' $\endgroup$ – xuce1234 Jun 3 '19 at 10:38
  • $\begingroup$ How to find the closed form for general series $(p>3)$ ? - If by absurd such a closed form in terms of $\zeta$ functions were to be known to humanity, then it would almost automatically follow that the $\zeta$ function is irrational for all natural values of the argument, which is still an unproven conjecture. $($Basically, by replacing $4$ with $-1,$ we get the celebrated series for $\zeta(3),$ which Apery famously used to prove its irrationality$).$ $\endgroup$ – Lucian Jul 14 '19 at 4:33
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Note that $$\sum_{k=1}^\infty \frac{(4x)^n}{n^2{{2n}\choose n}}=2\arcsin^2(\sqrt{x}).$$ Hence, for $p=3$ we have the integral form $$\sum\limits\limits_{n=1}^\infty \frac{4^n}{n^3\binom{2n}{n}}=\int_{0}^1\frac{2\arcsin^2(\sqrt{x})}{x}\,dx.$$ and you should be able to recover the result $\pi^2\ln(2)-\frac{7}{2}\zeta(3)$.

As regards the case $p=4$, $$\sum_{n=1}^\infty \frac{4^n}{n^4\binom{2n}{n}}=\int_0^1\frac{1}{t}\int_{x=0}^t\frac{2\arcsin^2(\sqrt{x})}{x}\,dx\,dt$$ which, according to ykcaZ's comment below, leads to $$8\int_0^\frac{\pi}{2} x\ln^2(\sin x)dx$$ that is equal to $$8\operatorname{Li}_4\left(\frac{1}{2}\right)+\frac{1}{3}\ln^4(2)+\frac{2\pi^2}{3}\ln^2(2)-\frac{19\pi^4}{360}$$ (see tough definite integral: $\int_0^\frac{\pi}{2}x\ln^2(\sin x)~dx$ ).

More generally, for $p\geq 2$, $$\sum_{n=1}^\infty \frac{4^n}{n^p\binom{2n}{n}} =\frac{(-2)^p}{(p-2)!}\int_0^\frac{\pi}{2} x\ln^{p-2}(\sin x)\,dx.$$

Look through the paper Sums of reciprocals of the central binomial coefficients by R. Sprugnoli for more references. See also On binomial sums $\sum_{n=1}^\infty \frac{1}{n^k\,\binom {2n}n}$ and log sine integrals

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  • $\begingroup$ Thanks for the link and $\to +1$ $\endgroup$ – Claude Leibovici Jun 3 '19 at 11:06
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    $\begingroup$ For $p=4$ we easily get $$\sum_{n=1}^\infty \frac{4^n}{n^4\binom{2n}{n}}=-8\int_0^\frac{\pi}{2} x\ln^2(\sin x)dx$$ So I can bet there will be a closed form. $\endgroup$ – Zacky Jun 3 '19 at 11:51
  • $\begingroup$ So we have some kind of closed form. See my edit. $\endgroup$ – Robert Z Jun 3 '19 at 12:24
  • $\begingroup$ @RobertZ thanks for the link, the answer looks decent as I expected much worse. It's quite over my league already as I stared at the integral for like half an hour without having any idea on how to start. I wouldn't even think about $p=5$, but maybe a computer can deal with that too. $\endgroup$ – Zacky Jun 3 '19 at 12:27
  • $\begingroup$ You may like this related post. $\endgroup$ – Tito Piezas III Jun 30 '19 at 15:18
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We can make use of the following representation $$\sf 2\arcsin^2z=\sum\limits_{n\geq1}\frac {(2z)^{2n}}{n^2\binom {2n}n}, \ z\in[-1,1]$$ Which gives integrating once with respect to $\sf z$ from $\sf 0$ to $\sf x$: $$\sf 4\int_0^x \frac{\arcsin^2 z}{z}dz =\sum_{n=1}^\infty \frac{(2x)^{2n}}{n^3 \binom{2n}{n}}$$ So the sum can be written as $$\sf S_3=\sum_{n=1}^\infty \frac{4^n}{n^3\binom{2n}{n}}=4\int_0^1 \frac{\arcsin^2 t}{t}dt$$ Now let $\sf t=\sin x$ and integrate by parts in order to get: $$\sf S_3=4\int_0^\frac{\pi}{2} x^2\cot xdx=-8\int_0^\frac{\pi}{2} x\ln(\sin x)dx$$ Also we can use the fourier series of log sine $$\sf S_3=8\ln 2 \int_0^\frac{\pi}{2} xdx+8\sum_{n=1}^\infty \frac{1}{n}\int_0^\frac{\pi}{2}x\cos(2nx)dx$$ The second integral is easily doable integrating by parts, thus: $$\sf S_3=\pi^2 \ln 2+2\sum_{n=1}^\infty \frac{(-1)^n-1}{n^3} = \boxed{\pi^2\ln 2 -\frac72\zeta(3)}$$


For higher $p$ things will get quite complicate, but the approach is the same. For case $p=4$ we have: $$\sf \frac{4}{x}\int_0^x \frac{\arcsin^2 z}{z}dz =\sum_{n=1}^\infty \frac{4^{n}x^{2n-1}}{n^3 \binom{2n}{n}}$$ And integrating once again produces $$\sf 8\int_0^t\frac{1}{x}\int_0^x \frac{\arcsin^2 z}{z}dzdx =\sum_{n=1}^\infty \frac{4^{n}t^{2n}}{n^4 \binom{2n}{n}}$$ $$\sf \Rightarrow S_4=\sum_{n=1}^\infty \frac{4^{n}}{n^4 \binom{2n}{n}}=8\int_0^1\frac{1}{x}\int_0^x \frac{\arcsin^2 z}{z}dzdx$$ $$\sf =8\int_0^1\int_z^1 \frac{1}{x}\frac{\arcsin^2 z}{z}dx dz=-8\int_0^1 \frac{\arcsin^2 z \ln z}{z}dz$$ Set $z=\sin x$ and integrate by parts to get $$\sf S_4=-8\int_0^\frac{\pi}{2} x^2 \ln(\sin x)\cot x dx=8\int_0^\frac{\pi}{2} x\ln^2(\sin x)dx$$ $$=\boxed{8\operatorname{Li}_2\left(\frac12\right)+\frac13\ln^42 +4\zeta(2)\ln^2 2-\frac{19}{4}\zeta(4)}$$ See here for the above integral.


Or for $p=5$ we have by the same approach: $$\sf 8\int_0^y\frac{1}{t}\int_0^t\frac{1}{x}\int_0^x \frac{\arcsin^2 z}{z}dzdxdt =\sum_{n=1}^\infty \frac{4^{n}y^{2n}}{n^5 \binom{2n}{n}}$$ $$\sf \sum_{n=1}^\infty \frac{4^{n}}{n^5 \binom{2n}{n}}=8\int_0^1 \int_z^1\int_z^1 \frac{\arcsin^2 z}{xtz}dxdtdz=8\int_0^1 \frac{\arcsin^2 z\ln^2 z}{z}dz$$ $$\sf \overset{z=\sin x}=8\int_0^\frac{\pi}{2}x^2\ln^2(\sin x)\cot x dx \overset{IBP}=-\frac{16}3\int_0^\frac{\pi}{2} x\ln^3(\sin x)dx$$ Furthermore this paper may be useful.

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    $\begingroup$ (+1) Nice answer. $\endgroup$ – Robert Z Jun 3 '19 at 13:13
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Setting $$n^{-p} = \frac{1}{\Gamma(p)} \int_{0}^{\infty} t^{p-1} e^{- n t}$$

I find for the sum

$$s(p) = \zeta(p) + \frac{1}{\Gamma(p)} \int_0^\infty t^{p-1}\frac{e^{- t/2} }{( 1-e^{-t} )^{\frac{3}{2}}} \arcsin(e^{-t/2})$$

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