A nice challenge by Cornel Valean:
Show that
$$2\sum _{n=1}^{\infty }\frac{2^{4 n}}{\displaystyle n^3 \binom{2 n}{n}^2}-\sum _{n=1}^{\infty }\frac{2^{4 n}}{\displaystyle n^4 \binom{2 n}{n}^2}+\sum _{n=1}^{\infty }\frac{2^{4 n} H_n^{(2)}}{\displaystyle n^2 (2 n+1) \binom{2 n}{n}^2}=\frac{\pi^3}{3}.$$
I have to say that I am not experienced in series involving squared central binomial coefficient, so I leave it for people who are experts in such series.
All approaches are appreciated. Thank you.
This is a possible approach. The sum of the OP can be rewritten as
$$I=\sum _{n=1}^{\infty }\frac{2n\cdot 2^{4 n}}{\displaystyle n^4\binom{2 n}{n}^2}-\sum _{n=1}^{\infty }\frac{2^{4 n}}{\displaystyle n^4 \binom{2 n}{n}^2}+\sum _{n=1}^{\infty }\frac{n^2\, 2^{4 n} H_n^{(2)}}{\displaystyle n^4 (2 n+1) \binom{2 n}{n}^2}\\ = \sum _{n=1}^{\infty }\frac{2^{4 n}[(2n-1)(2n+1)+n^2H_n^{(2)}]}{\displaystyle n^2 (2 n+1) \binom{2 n}{n}^2} \\ = \sum _{n=1}^{\infty }\frac{2^{4 n}(4n^2-1+n^2H_n^{(2)})}{\displaystyle n^4 (2 n+1) \binom{2 n}{n}^2} \\ = \sum _{n=1}^{\infty }\frac{2^{4 n}(4-1/n^2+H_n^{(2)})}{\displaystyle n^2 (2 n+1) \binom{2 n}{n}^2} \\ = \sum _{n=1}^{\infty }\frac{2^{4 n}(4+H_{n-1}^{(2)})}{\displaystyle n^2 (2 n+1) \binom{2 n}{n}^2} \\ = \sum _{n=1}^{\infty }\frac{2^{4 n} \, (n!)^4\, (4+H_{n-1}^{(2)})}{\displaystyle n^2(2 n+1)(2n!)^2} \\ = \sum _{n=1}^{\infty }\frac{ (2n!!)^4\, (4+H_{n-1}^{(2)})}{\displaystyle n^2(2 n+1)(2n!)^2} \\ = \sum _{n=1}^{\infty }\frac{ (2n!!)^2\, (4+H_{n-1}^{(2)})}{\displaystyle n^2 (2 n+1)(2n-1!!)^2} \\ = 4 \,\, \underbrace{ \sum _{n=1}^{\infty } \frac{ (2n!!)^2}{\displaystyle n^2 (2 n+1)(2n-1!!)^2} }_\text{J} \\ +\underbrace{\sum _{n=1}^{\infty } \sum_{k=1}^{n-1} \frac{ (2n!!)^2}{\displaystyle n^2 (2 n+1)(2n-1!!)^2} \frac{1}{k^2}}_\text{K} \\ $$
So we have $I=4J+K$. Let us consider firstly the terms of the $J$ summation. For given $n$, the corresponding summand $j_n$ is given by
$$j_n=\frac{1}{n^2} \prod_{k=1}^n \frac{4 k^2}{(2 k- 1) (2 k + 1)} \\=\frac{1}{n^2}\left(\frac 21\cdot \frac 23 \right)\cdot \left(\frac 43\cdot \frac 45 \right)... \cdot \left(\frac{2n}{2n-1}\cdot \frac {2n}{2n+1}\right)$$
where the infinite product resembles the classical Wallis formula for $\pi/2$. Note that the terms satisfy the recurrence
$$j_{n+1}=j_n \frac{n^2(2n+2)^2}{(n+1)^2(2n+1)(2n+3)}\\ = j_n \frac{4n^2}{(2n+1)(2n+3)} $$
and that they can be written in the form
$$j_n=\frac{\pi \,Γ^2(n + 1)}{2 n^2\, Γ(n + \frac 12) Γ(n + \frac 32)}$$
Moreover, the terms in the $J$ summation satisfies the interesting property
$$\sum_{n=1}^m j_n=4m^2 j_m-4$$
We will prove it by induction. For $m=1$, the sum reduces to the single term $j_1=4/3$, and accordingly $4 \cdot 1^2\cdot 4/3-4=4/3$. Now let us assume that the property is valid for a given $m$. Passing to $m+1$, the sum becomes
$$\sum_{n=1}^{m+1} j_n=4m^2 j_m-4 +j_{m+1}\\ =4m^2 \frac{(2m+1)(2m+3)} {m^2}j_{m+1}-4+j_{m+1} \\ = (4n^2+8n+3) j_{m+1}-4+j_{m+1}\\ = (4n^2+8n+4) j_{m+1}-4 \\ = 4(m+1)^2 j_{m+1}-4 $$
so that the property is still valid for $m+1$, and the claim is proved. Then we have
$$\sum_{n=1}^m j_n= \frac{2\pi \,\Gamma^2(n + 1)}{ \Gamma(n + \frac 12 ) \Gamma(n + \frac 32 )}-4 $$
Taking the limit for $m\rightarrow \infty$, since
$$\lim_{m\rightarrow \infty} \frac{\Gamma^2(m + 1)}{ \Gamma(m + \frac 12 ) \Gamma(m + \frac 32 )}=1$$
we have
$$J= \sum_{n=1}^\infty j_n = 2\pi-4$$
in accordance with the numerical approximation of $J \approx 2.283$ given by WA here.
For the double summation $K$, writing it again in terms of Gamma functions as already done for $J$, using the same definition of $j_n$ given above, and swapping the indices we have
$$K=\sum _{k=1}^{\infty } \sum_{n=k+1}^{\infty} j_n \cdot \frac{1}{k^2}\\ =\sum _{k=1}^{\infty } \frac{1}{k^2} \sum_{n=1}^{\infty} j_n - \sum _{k=1}^{\infty } \frac{1}{k^2} \sum_{n=1}^{k} j_n \\ = \frac{\pi^2}{6} (2\pi-4) -\sum _{k=1}^{\infty } \frac{1}{k^2} \left[ \frac{2\pi \,\Gamma^2(k + 1)}{ \Gamma(k + \frac 12 ) \Gamma(k + \frac 32 )}-4\right] \\ = \frac{\pi^3}{3} - \frac{2\pi^2}{3}+ 4\sum _{k=1}^{\infty } \frac{1}{k^2} \\ -\sum _{k=1}^{\infty } \left[ \frac{2\pi \,\Gamma^2(k + 1)}{ k^2\Gamma(k + \frac 12 ) \Gamma(k + \frac 32 )}\right] $$
and since the last summation is equivalent to $4\sum_{n=1}^\infty j_n$,
$$K=\frac{\pi^3}{3} -4(2\pi-4)$$
We then conclude that
$$I=4J+K \\=4(2\pi-4) + \frac{\pi^3}{3} -4(2\pi-4) = \frac{\pi^3}{3}$$
Here's a simple way to prove (with the manipulation of well-known generating functions and integrals) that:
$$S=2\sum _{k=1}^{\infty }\frac{16^k}{k^3\binom{2k}{k}^2}-\sum _{k=1}^{\infty }\frac{16^k}{k^4\binom{2k}{k}^2}+\sum _{k=1}^{\infty }\frac{16^kH_k^{\left(2\right)}}{k^2\left(2k+1\right)\binom{2k}{k}^2}=\frac{\pi ^3}{3}.$$
First let's make use of the following generating function: $$\sum _{k=1}^{\infty }\frac{4^kx^{2k}}{k^2\binom{2k}{k}}=2\arcsin ^2\left(x\right)$$ $$\sum _{k=1}^{\infty }\frac{4^ky^{2k}}{k^3\binom{2k}{k}}=4\int _0^y\frac{\arcsin ^2\left(x\right)}{x}\:dx$$ $$2\sum _{k=1}^{\infty }\frac{4^k}{k^3\binom{2k}{k}}\int _0^{\frac{\pi }{2}}\sin ^{2k-1}\left(y\right)\:dy=8\int _0^{\frac{\pi }{2}}\csc \left(y\right)\int _0^{\sin \left(y\right)}\frac{\arcsin ^2\left(x\right)}{x}\:dx\:dy$$ $$\boxed{\sum _{k=1}^{\infty }\frac{16^k}{k^4\binom{2k}{k}^2}=8\int _0^{\frac{\pi }{2}}\csc \left(y\right)\int _0^yx^2\cot \left(x\right)\:dx\:dy}.$$ Now let's jump into the $3$rd series by first considering the same generating function used previously: $$\sum _{k=1}^{\infty }\frac{4^kx^{2k}}{k^2\binom{2k}{k}}=2\arcsin ^2\left(x\right)$$ $$\sum _{k=1}^{\infty }\frac{4^k}{k^3\left(2k+1\right)\binom{2k}{k}}t^{2k}=4\int _0^t\left(2\frac{\sqrt{1-y^2}\arcsin \left(y\right)}{y^2}-2\frac{1}{y}+\frac{\arcsin ^2\left(y\right)}{y}\right)\:dy$$ $$\sum _{k=1}^{\infty }\frac{16^k}{k^4\left(2k+1\right)\binom{2k}{k}^2}=8\int _0^{\frac{\pi }{2}}\csc \left(t\right)\int _0^t\left(2x\cot ^2\left(x\right)-2\cot \left(x\right)+x^2\cot \left(x\right)\right)\:dx\:dt$$ $$\boxed{\sum _{k=1}^{\infty }\frac{16^k}{k^4\left(2k+1\right)\binom{2k}{k}^2}=16\int _0^{\frac{\pi }{2}}\csc \left(t\right)\left(1-t\cot \left(t\right)\right)\:dt-8\int _0^{\frac{\pi }{2}}t^2\csc \left(t\right)\:dt+\sum _{k=1}^{\infty }\frac{16^k}{k^4\binom{2k}{k}^2}}.$$ Lastly consider: $$\sum _{k=1}^{\infty }\frac{4^kH_k^{\left(2\right)}}{k\binom{2k}{k}}x^{2k}-\sum _{k=1}^{\infty }\frac{4^k}{k^3\binom{2k}{k}}x^{2k}=\frac{4}{3}\frac{x\arcsin ^3\left(x\right)}{\sqrt{1-x^2}}$$ $$\sum _{k=1}^{\infty }\frac{4^kH_k^{\left(2\right)}}{k\left(2k+1\right)\binom{2k}{k}}y^{2k+1}-\sum _{k=1}^{\infty }\frac{4^k}{k^3\left(2k+1\right)\binom{2k}{k}}y^{2k+1}=\frac{4}{3}\int _0^y\frac{x\arcsin ^3\left(x\right)}{\sqrt{1-x^2}}\:dx$$ $$\sum _{k=1}^{\infty }\frac{16^kH_k^{\left(2\right)}}{k^2\left(2k+1\right)\binom{2k}{k}^2}-\sum _{k=1}^{\infty }\frac{16^k}{k^4\left(2k+1\right)\binom{2k}{k}^2}=\frac{8}{3}\int _0^{\frac{\pi }{2}}\csc ^2\left(y\right)\int _0^{\sin \left(y\right)}\frac{x\arcsin ^3\left(x\right)}{\sqrt{1-x^2}}\:dx\:dy$$ $$\boxed{\sum _{k=1}^{\infty }\frac{16^kH_k^{\left(2\right)}}{k^2\left(2k+1\right)\binom{2k}{k}^2}=\frac{8}{3}\int _0^{\frac{\pi }{2}}x^3\cos \left(x\right)\:dx+\sum _{k=1}^{\infty }\frac{16^k}{k^4\left(2k+1\right)\binom{2k}{k}^2}}.$$ And finally from here we know that: $$\sum _{k=1}^{\infty }\frac{16^k}{k^3\binom{2k}{k}^2}=4\int _0^{\frac{\pi }{2}}x^2\csc \left(x\right)\:dx,$$ plugging these results into the main expression we obtain: $$S=\require{cancel}\cancel{8\int _0^{\frac{\pi }{2}}x^2\csc \left(x\right)\:dx}-\cancel{\sum _{k=1}^{\infty }\frac{16^k}{k^4\binom{2k}{k}^2}}+\frac{8}{3}\int _0^{\frac{\pi }{2}}x^3\cos \left(x\right)\:dx$$ $$\require{cancel}+16\int _0^{\frac{\pi }{2}}\csc \left(t\right)\left(1-t\cot \left(t\right)\right)\:dt-\cancel{8\int _0^{\frac{\pi }{2}}t^2\csc \left(t\right)\:dt}+\cancel{\sum _{k=1}^{\infty }\frac{16^k}{k^4\binom{2k}{k}^2}}.$$ Therefore: $$S=2\sum _{k=1}^{\infty }\frac{16^k}{k^3\binom{2k}{k}^2}-\sum _{k=1}^{\infty }\frac{16^k}{k^4\binom{2k}{k}^2}+\sum _{k=1}^{\infty }\frac{16^kH_k^{\left(2\right)}}{k^2\left(2k+1\right)\binom{2k}{k}^2}$$ $$=\frac{8}{3}\int _0^{\frac{\pi }{2}}x^3\cos \left(x\right)\:dx+16\int _0^{\frac{\pi }{2}}\csc \left(t\right)\left(1-t\cot \left(t\right)\right)\:dt=\frac{\pi ^3}{3},$$ and that is it.
An excellent answer was already given (the chosen one), but good to have more ways in place.
A solution by Cornel Ioan Valean
Instead of calculating all three series separately, we might try to calculate them all at once. So, we have that $$2\sum _{n=1}^{\infty }\frac{2^{4 n}}{\displaystyle n^3 \binom{2 n}{n}^2}-\sum _{n=1}^{\infty }\frac{2^{4 n}}{\displaystyle n^4 \binom{2 n}{n}^2}+\sum _{n=1}^{\infty }\frac{2^{4 n} H_n^{(2)}}{\displaystyle n^2 (2 n+1) \binom{2 n}{n}^2}$$ $$=\sum _{n=1}^{\infty }\frac{2^{4n} (4n^2-1+n^2 H_n^{(2)})}{\displaystyle n^4 (2 n+1) \binom{2 n}{n}^2}=\sum _{n=1}^{\infty }\frac{2^{4n} (4-1/n^2+ H_n^{(2)})}{\displaystyle n^2 (2 n+1) \binom{2 n}{n}^2}$$ $$=\sum _{n=1}^{\infty }\frac{2^{4n}(4-1/n^2+ H_n^{(2)}\color{blue}{+(4 n^2-1) H_{n-1}^{(2)}}-\color{blue}{(4 n^2-1) H_{n-1}^{(2)}})}{\displaystyle n^2 (2 n+1) \binom{2 n}{n}^2}$$ $$=\sum _{n=1}^{\infty }\frac{2^{4n}(\color{red}{4n^2H_n^{(2)}}-\color{blue}{(4 n^2-1) H_{n-1}^{(2)}})}{\displaystyle n^2 (2 n+1) \binom{2 n}{n}^2}$$ $$=\sum _{n=1}^{\infty}\left(\frac{2^{4n+2}H_n^{(2)}}{\displaystyle (2n+1) \binom{2 n}{n}^2}-\frac{2^{4n}(2n-1)H_{n-1}^{(2)} }{\displaystyle n^2\binom{2 n}{n}^2}\right)$$ $$=\lim_{N\to\infty}\sum _{n=1}^{N}\left(\frac{2^{4n+3}H_n^{(2)}}{\displaystyle (n+1) \binom{2 n+2}{n+1}\binom{2 n}{n}}-\frac{2^{4n-1}H_{n-1}^{(2)} }{\displaystyle n\binom{2 n}{n}\binom{2 n-2}{n-1}}\right)$$ $$=\lim_{N\to\infty}\frac{2^{4N+3}H_N^{(2)}}{\displaystyle (N+1) \binom{2 N+2}{N+1}\binom{2 N}{N}}=\frac{\pi^3}{3},$$
where we used the asymptotic form of the central binomial coefficient, $\displaystyle \binom{2 N}{N}\sim \frac{4^N}{\sqrt{\pi N}}$.
