I want to know what strategies I can use in order to evaluate: $$\sum _{n=1}^{\infty }\frac{H_{2n}^2\binom{2n}{n}}{4^nn^2}.$$ What I tried is the following: $$2\int _0^1x^{2n-1}\ln ^2\left(1-x\right)\:\mathrm{d}x=\frac{H_{2n}^2+H_{2n}^{\left(2\right)}}{n}$$ $$2\int _0^1\left(\sum _{n=1}^{\infty }\frac{\binom{2n}{n}}{4^nn}x^{2n}\right)\frac{\ln ^2\left(1-x\right)}{x}\:\mathrm{d}x=\sum _{n=1}^{\infty }\frac{H_{2n}^2\binom{2n}{n}}{4^nn^2}+\sum _{n=1}^{\infty }\frac{H_{2n}^{\left(2\right)}\binom{2n}{n}}{4^nn^2}$$ $$\sum _{n=1}^{\infty \:}\frac{H_{2n}^2\binom{2n}{n}}{4^nn^2}=-4\int _0^1\frac{\ln ^2\left(1-x\right)\ln \left(1+\sqrt{1-x^2}\right)}{x}\:\mathrm{d}x+4\ln \left(2\right)\int _0^1\frac{\ln ^2\left(1-x\right)}{x}\:\mathrm{d}x-\sum _{n=1}^{\infty }\frac{H_{2n}^{\left(2\right)}\binom{2n}{n}}{4^nn^2},$$ but I got stuck trying to evaluate: $$\int _0^1\frac{\ln ^2\left(1-x\right)\ln \left(1+\sqrt{1-x^2}\right)}{x}\:\mathrm{d}x.$$ Any other ideas to evaluate the main sum or perhaps that integral?
2 Answers
Such sums can be evaluated pretty much systematically via MZV, giving $$\DeclareMathOperator{Li}{Li} \color{royalblue}{8 \pi \Im\left(\text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)+24 \text{Li}_4\left(\frac{1}{2}\right)+\frac{7}{4} \zeta (3) \log (2)-\frac{149 \pi ^4}{720}+\log ^4(2)-\frac{5}{4} \pi ^2 \log ^2(2)} \approx 2.35042$$
Note that $\int_0^1 \frac{x^n}{\sqrt{x(1-x)}} dx = \pi \binom{2n}{n}4^{-n}$, so for $f(x) = \sum_{n\geq 1} \frac{H_{2n}^2}{n^2}x^n$, the sum is $$S = \frac{1}{\pi} \int_0^1 \frac{f(x)}{\sqrt{x(1-x)}}dx = \frac{2}{\pi} \int_0^1 \frac{f(x^2)}{\sqrt{1-x^2}}dx$$ write $f(x^2) = 2g(x) + 2g(-x)$ with $g(x) = \sum_{n\geq 1} \frac{H_n^2}{n^2}x^n$, $$\tag{*} S = \frac{4}{\pi} \int_{-1}^1 \frac{g(x)}{\sqrt{1-x^2}} dx = \frac{4}{\pi} \int_{-1}^1 g(\frac{2t}{1+t^2}) \frac{2}{1+t^2} dt$$
Recall the generalized polylogarithm $$\Li_{s_1,\cdots,s_d}(x) = \sum_{n_1 > n_2 > \cdots > n_d \geq 1} \frac{x^{n_1}}{n_1^{s_1} n_2^{s_2} \cdots n_d^{s_d}}$$ $g(x)$ is a weight $4$ generalized polylog, explicitly, $$g(x) = \Li_4(x) + 2\Li_{3,1}(x) + 2\Li_{2,1,1}(x) + \Li_{2,2}(x)$$
It's a result that integrals like those in (*) are level 4 CMZVs, and there is a routine algorithm to convert them into CMZVs, expressing them on a convenient $\mathbb{Q}$-basis gives the result.
A Mathematica package introduced here automatically performs these steps:
There are many variations, which can be solved using same approach (as long as weight and level is low enough), for example:
$$\sum _{n=1}^{\infty }\frac{H_{2n}H^{(2)}_{2n}}{n} \frac{1}{a_n} \qquad \sum _{n=1}^{\infty }\frac{(H^{(2)}_{n})^2}{n^2} a_n \qquad \sum _{n=1}^{\infty }\frac{H_{n}^5}{n} \frac{1}{a_n} \qquad a_n = 4^n \binom{2n}{n}^{-1}$$
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A simple idea: to make the problem accessible by simple means all you have to do is to denote $\displaystyle I=\int _0^1\frac{\log^2\left(1-x\right)\log \left(1+\sqrt{1-x^2}\right)}{x}\mathrm{d}x$ and $\displaystyle J=\int _0^1\frac{\log^2\left(1-x\right)\log \left(1-\sqrt{1-x^2}\right)}{x}\mathrm{d}x$, and then consider $I+J$ and $I-J$, which will immediately reveal (by some substitutions) known integrals, or integrals that are easily reducible to manageable harmonic series (of course, in some cases we can also avoid the use of harmonic series).
That's all (indeed, a bit of boring work).
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