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My question is at the end of the solution.

We have

$$\int_0^1 x^{n-1}\ln(1-x)=-\frac{H_n}{n}$$

Differentiate both sides with respect to $n$

$$\int_0^1 x^{n-1}\ln x\ln(1-x)=\frac{H_n}{n^2}+\frac{H_n^{(2)}}{n}-\frac{\zeta(2)}{n}$$

Next multiply both sides by $\frac{4n}{n{2n\choose n}}$ then $\sum_{n=1}^\infty$ we get

$$\sum_{n=1}^\infty\frac{4^nH_n}{n^3{2n\choose n}}+\sum_{n=1}^\infty\frac{4^nH_n^{(2)}}{n^2{2n\choose n}}-\zeta(2)\sum_{n=1}^\infty\frac{4^n}{n^2{2n\choose n}}=\int_0^1\frac{\ln x\ln(1-x)}{x}\left(\sum_{n=1}^\infty\frac{(4x)^n}{n{2n\choose n}}\right)dx$$

$$=\int_0^1\frac{\ln x\ln(1-x)}{x}\left(\frac{2\sqrt{x}\arcsin\sqrt{x}}{\sqrt{1-x}}\right)dx$$

$$\overset{\sqrt{x}=\sin\theta}{=}16\int_0^{\pi/2}\theta\ln(\sin\theta)\ln(\cos\theta)d\theta=16I$$

For the integral, let $\theta\to \frac{\pi}{2}-\theta$ to have

$$I=\frac{\pi}{2}\int_0^{\pi/2}\ln(\sin\theta)\ln(\cos\theta)d\theta-\int_0^{\pi/2}\theta\ln(\sin\theta)\ln(\cos\theta)d\theta$$

$$\Longrightarrow 2I=\frac{\pi}{2}\int_0^{\pi/2}\ln(\sin\theta)\ln(\cos\theta)d\theta$$

$$=\frac{\pi}{2}\left(\frac{\pi}{2}\ln^2(2)-\frac{\pi^3}{48}\right)$$

$$\Longrightarrow I=\frac34\ln^2(2)\zeta(2)-\frac{15}{32}\zeta(4)$$

Therefore

$$\sum_{n=1}^\infty\frac{4^nH_n}{n^3{2n\choose n}}+\sum_{n=1}^\infty\frac{4^nH_n^{(2)}}{n^2{2n\choose n}}-\zeta(2)\sum_{n=1}^\infty\frac{4^n}{n^2{2n\choose n}}=12\ln^2(2)\zeta(2)-\frac{15}{2}\zeta(4)$$

Since

$$\zeta(2)\sum_{n=1}^\infty\frac{4^n}{n^2{2n\choose n}}=\zeta(2)\left(\frac{\pi^2}{2}\right)=\frac{15}{2}\zeta(4)$$

we have the nice relation

$$\sum_{n=1}^\infty\frac{4^nH_n}{n^3{2n\choose n}}+\sum_{n=1}^\infty\frac{4^nH_n^{(2)}}{n^2{2n\choose n}}=12\ln^2(2)\zeta(2)$$

Finally substitute

$$\sum_{n=1}^\infty\frac{4^nH_n}{n^3{2n\choose n}}=-8\text{Li}_4\left(\frac12\right)+\zeta(4)+8\ln^2(2)\zeta(2)-\frac{1}{3}\ln^4(2)$$

we obtain

$$\sum_{n=1}^\infty\frac{4^nH_n^{(2)}}{n^2{2n\choose n}}=8\text{Li}_4\left(\frac12\right)-\zeta(4)+4\ln^2(2)\zeta(2)+\frac{1}{3}\ln^4(2)\approx 6.2957$$

But Mathematica gives

$$\sum_{n=1}^\infty\frac{4^nH_n^{(2)}}{n^2{2n\choose n}}\approx 6.04326$$

Can you spot any mistake or my solution is good?

Thank you.

Ali Shadhar
  • 25,498
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    A bonus one: $$\sum _{n=1}^{\infty } \frac{4^n \left(H_n\right){}^3}{n^2 \binom{2 n}{n}}=80 \text{Li}_5\left(\frac{1}{2}\right)-\frac{155 \zeta (5)}{8}+3 \pi ^2 \zeta (3)-\frac{2 \log ^5(2)}{3}+\frac{16}{9} \pi ^2 \log ^3(2)+\frac{91}{36} \pi ^4 \log (2)$$ Have fun! – Infiniticism Aug 17 '20 at 06:06

1 Answers1

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@User 628759 gave the solution. It is quite surprising to see how sensitive it the result to this parameter.

On my side, I computed exactly $$\sum_{n=1}^{10000}\frac{4^nH_n^{(2)}}{n^2{2n\choose n}}\approx 6.23740$$ $$\sum_{n=1}^{20000}\frac{4^nH_n^{(2)}}{n^2{2n\choose n}}\approx 6.25448$$

To be slow, it is !

Claude Leibovici
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