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.
On
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, $\displaystyle \binom{2 N}{N}\sim \frac{4^N}{\sqrt{\pi N}}$.
On
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.
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
$$I=4J+K \\=4(2\pi-4) + \frac{\pi^3}{3} -4(2\pi-4) = \frac{\pi^3}{3}$$