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I'm wondering if one could derive a closed form expression for the series $$\sum_{n=1}^{\infty}\frac{H_{2n}}{n^2\cdot4^n}{2n \choose n}$$

$$\text{With } \text{ } \text{ } \text{ }H_n=\sum_{k=1}^{n}\frac{1}{k}\text{ } \text{ } \text{} \text{ } \text{ }\text{the } n^{th} \text{ harmonic number.}$$

Now, I know series involving harmonic numbers are well suited for a summation by part (or Abel's transformation) approach, but it doesn't lead anywere here, at least not in this state.

Any suggestions ?

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  • $\begingroup$ Don't you mean $H_n=\sum_{k=1}^{n}\frac{1}{k}$? $\endgroup$ Jan 24, 2019 at 1:13
  • $\begingroup$ Oh yes sorry, now it's corrected ! $\endgroup$ Jan 24, 2019 at 1:14
  • $\begingroup$ Would the fact that $\frac{{2n\choose n}}{4^n}=\frac{1}{2\pi}\int_0^{2\pi} \cos^{2n} x\,dx$ be useful if you can switch integration and summation? $\endgroup$
    – Dispersion
    Jan 24, 2019 at 1:47
  • $\begingroup$ Looks like a very good idea, but as far as I know there is no known closed form for the power series $\sum\limits_{n=1}^{\infty}\frac{H_{2n}}{n^2}x^{2n}$... And even if there were, I guess we would eventually get a very nasty looking integral, especially after composing with the cosine function... $\endgroup$ Jan 24, 2019 at 2:32

4 Answers 4

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For $x \in [0,1]$ let $$ f(x) = \sum \limits_{n=1}^\infty \frac{{2n \choose n}}{n^2 4^n} x^{2n} \, . $$ Using the power series of $\arcsin$ we find $$ x \frac{\mathrm{d}}{\mathrm{d} x} x \frac{\mathrm{d}}{\mathrm{d} x} f(x) = 4 \frac{\mathrm{d}}{\mathrm{d} x} [\arcsin(x) - x] = 4 \left[\frac{1}{\sqrt{1-x^2}} - 1 \right] $$ for $x \in [0,1)$ . In particular, $$ f'(1) = 4 \int \limits_0^1 \frac{1}{x} \left[\frac{1}{\sqrt{1-x^2}} - 1 \right] \, \mathrm{d} x \stackrel{x=\sqrt{1-y^2}}{=} 4 \int \limits_0^1 \frac{\mathrm{d} y}{1+y} = 4 \ln(2) \, . $$ Now we can compute \begin{align} S &\equiv \sum \limits_{n=1}^\infty \frac{H_{2n} {2n \choose n}}{n^2 4^n} = \sum \limits_{n=1}^\infty \frac{{2n \choose n}}{n^2 4^n} \int \limits_0^1 \frac{1-x^{2n}}{1-x} \, \mathrm{d} x = \int \limits_0^1 \frac{f(1) - f(x)}{1-x} \, \mathrm{d} x \\ &= \int \limits_0^1 \frac{- \ln(1-x)}{x} x f'(x) \, \mathrm{d} x = \operatorname{Li}_2 (1) f'(1) - 4 \int \limits_0^1 \frac{\operatorname{Li}_2 (x)}{x} \left[\frac{1}{\sqrt{1-x^2}} - 1 \right] \, \mathrm{d} x \\ &= \operatorname{Li}_2 (1) f'(1) + 4 \operatorname{Li}_3 (1) - 4 \int \limits_0^1 \frac{\operatorname{Li}_2 (x)}{x \sqrt{1-x^2}} \, \mathrm{d} x \equiv 4 \left[\frac{\pi^2}{6} \ln(2) + \zeta(3) - I\right] \, . \end{align} In order to find $I$ we use a well-known integral representation of the dilogarithm: \begin{align} I &= \int \limits_0^\infty t \int \limits_0^1 \frac{\mathrm{d} x}{(\mathrm{e}^t - x) \sqrt{1-x^2}} \, \mathrm{d} t \stackrel{(*)}{=} \int \limits_0^\infty \frac{t \left[\frac{\pi}{2} + \arcsin(\mathrm{e}^{-t})\right]}{\sqrt{\mathrm{e}^{2t}-1}} \, \mathrm{d} t \\ &\stackrel{\mathrm{e}^{-t} = \sin(u)}{=} \frac{1}{2} \int \limits_0^{\pi/2} -\ln[\sin(u)] (\pi + 2 u) \, \mathrm{d} u = \frac{1}{2} \int \limits_0^{\pi/2} u (\pi + u) \cot(u) \, \mathrm{d} u \\ &= \frac{1}{2} [\pi K_1^{(1)} + K_2^{(1)}] = \frac{3}{8}\pi^2 \ln(2) - \frac{7}{16} \zeta(3) \, . \end{align} The integrals $ K_n^{(m)}$ are discussed in this question. Combining this result and the previous expression for the sum we end up with $$ \boxed{S = \sum \limits_{n=1}^\infty \frac{H_{2n} {2n \choose n}}{n^2 4^n} = \frac{23}{4} \zeta(3) - \frac{5}{6} \pi^2 \ln(2)} \, . $$


Proof of $(*)$:

For $a \in [0,1]$ let $$ g(a) = \int \limits_0^1 \frac{-\ln(1-a x)}{x \sqrt{1-x^2}} \, \mathrm{d} x= \sum \limits_{n=1}^\infty \frac{a^n}{n} \int \limits_0^{\pi/2} \sin^{n-1} (t) \, \mathrm{d} t \, .$$ Using Wallis' integrals we find $$ g(a) = \frac{\pi}{2} \sum \limits_{k=0}^\infty \frac{{2k \choose k} a^{2k+1}}{4^k(2k+1)} + \frac{1}{4} \sum \limits_{m=1}^\infty \frac{4^k a^{2k}}{k^2 {2k \choose k}} = \frac{\pi}{2} \arcsin(a) + \frac{1}{2} \arcsin^2 (a) \, . $$ Therefore $$ \int \limits_0^1 \frac{\mathrm{d} x}{(1-a x)\sqrt{1-x^2}} = g'(a) = \frac{\frac{\pi}{2} + \arcsin{a}}{\sqrt{1-a^2}} $$ holds for $a \in [0,1)$ .

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    $\begingroup$ This is a nice answer. Now, when I look at my junk, I am just ashamed. $\endgroup$ Jan 24, 2019 at 5:44
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    $\begingroup$ Yes indeed, this is a very nice answer. $\endgroup$
    – omegadot
    Jan 24, 2019 at 5:46
  • $\begingroup$ Thank you very much, this is a great answer ! $\endgroup$ Jan 24, 2019 at 15:03
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This is not an answer but it is too long for a comment.

Considering $$a_n=\frac{H_{2n}}{n^2\,4^n}{2n \choose n}\qquad \text{and} \qquad S_p=\sum_{n=1}^p a_n$$ none of the CAS I tried was able to find an expression for the partial sums or the infinite sum. Numerically, as shown below, the convergence looks to be extremely slow $$\left( \begin{array}{cc} p & S_p \\ 1000 & 1.21081501745 \\ 2000 & 1.21088004598 \\ 3000 & 1.21089738494 \\ 4000 & 1.21090493158 \\ 5000 & 1.21090901996 \\ 6000 & 1.21091153294 \\ 7000 & 1.21091321066 \\ 8000 & 1.21091439815 \\ 9000 & 1.21091527609 \\ 10000 & 1.21091594745 \end{array} \right)$$ which can be explained by the fact that, for large values of $n$ $$\frac {a_{n+1}} {a_n} \simeq 1+\frac{2-5( \log (2n)+ \gamma) }{2 n \left(\log(2n)+\gamma \right)}$$ For the infinite summation, the result seems to be close to $1.2109201$ which is not identified by inverse symbolic calculators.

For large values of $n$, we also can find $$a_n\simeq b_n=\frac{ \log (2n)+\gamma }{n^{5/2}\,\sqrt{\pi }}$$ which does not help much even if $$\sum_{n=1}^\infty b_n=\frac{(\gamma+\log(2)) \zeta \left(\frac{5}{2}\right)-\zeta '\left(\frac{5}{2}\right)}{\sqrt{\pi }}\approx 1.18001$$ However, numerically, this can be of some help writing $$S_\infty=S_p+\sum_{n=p+1}^\infty b_n$$ $$\left( \begin{array}{cc} p & S_\infty \approx \\ 100 & 1.2109213325 \\ 200 & 1.2109203863 \\ 300 & 1.2109202368 \\ 400 & 1.2109201900 \\ 500 & 1.2109201700 \\ 600 & 1.2109201590 \\ 700 & 1.2109201535 \\ 800 & 1.2109201498 \\ 900 & 1.2109201475 \\ 1000 & 1.2109201458 \end{array} \right)$$

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  • $\begingroup$ First five terms summed of a partial closed form sum approximation achieved using Mathematica 11.3. After term 9 the $\log[2]$ term disappears and then appears occasionally after that, not sure if this is an error or feature of this partial series. $$\text{FullSimplify}\left[2 \sum _{m=1}^5 \left(H_{2 m}-H_{2 m-2}\right) \int_0^1 \text{FullSimplify}\left[\sum _{n=m}^{\infty } \frac{\binom{2 n}{n} x^{2 n-1}}{n 2^{2 n}}\right] \, dx\right]=\frac{11 \left(515328 \pi ^2-859301-6183936 \log ^2(2)\right)}{11612160}$$ $\endgroup$ Jan 24, 2019 at 12:39
  • $\begingroup$ or for slightly simpler integral $$\text{FullSimplify}\left[2 \sum _{m=1}^5 \left(H_{2 m}-H_{2 m-2}\right) \int_0^{\frac{1}{2}} \text{FullSimplify}\left[\sum _{n=m}^{\infty } \frac{\binom{2 n}{n} x^{2 n-1}}{n}\right] \, dx\right] $$ $\endgroup$ Jan 24, 2019 at 13:20
  • $\begingroup$ First comment should read after partial term 10 $\endgroup$ Jan 24, 2019 at 13:27
  • $\begingroup$ @JamesArathoon. Very interesting and I thank you. Would you mind to send me the input form MMA syntax in a .txt file . I should try to play with it when going to university (my e-mail address is my profile). Thanks again :-) $\endgroup$ Jan 24, 2019 at 16:00
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Using the fact that $$\int_0^1x^{2n-1}\ln(1-x)\ dx=-\frac{H_{2n}}{2n}$$ multiply both sides by $\ \displaystyle-\frac{2}{n4^n}{2n \choose n}$ then take the sum, we get, \begin{align} S&=\sum_{n=1}^\infty\frac{H_{2n}}{n^24^n}{2n \choose n}=-2\int_0^1\frac{\ln(1-x)}{x}\left(\sum_{n=1}^\infty\frac{(x^2)^n}{n4^n}{2n \choose n}\right)\ dx \end{align} I managed here to prove: $$\quad\displaystyle\sum_{n=1}^\infty \frac{x^n}{n4^n}{2n \choose n}=-2 \tanh^{-1}{\sqrt{1-x}}-\ln x+2\ln2$$ which follows: \begin{align} S&=4\underbrace{\int_0^1\frac{\ln(1-x)\tanh^{-1}{\sqrt{1-x^2}}}{x}\ dx}_{\text{IBP}}+4\int_0^1\frac{\ln(1-x)\ln x}{x}\ dx-4\ln2\int_0^1\frac{\ln(1-x)}{x}\ dx\\ &=-4\int_0^1\frac{\operatorname{Li}_2(x)}{x\sqrt{1-x^2}}\ dx+4(\zeta(3))-4\ln2(-\zeta(2))\\ &=-4\left(\frac{3}{8}\pi^2 \ln(2) - \frac{7}{16} \zeta(3)\right)+4\zeta(3)+\frac{2}{3}\pi^2\ln(2)\\ &\boxed{=\frac{23}4\zeta(3)-\frac{5}{6}\pi^2\ln2} \end{align}

Credit goes to ComplexYetTrivial for nicely evaluating $\ \displaystyle\int_0^1\frac{\operatorname{Li}_2(x)}{x\sqrt{1-x^2}}\ dx=\frac{3}{8}\pi^2 \ln(2) - \frac{7}{16} \zeta(3)$

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Here is my way of evaluating this sum also offering a different way to calculate that polylogarithmic integral.

$$\sum _{k=1}^{\infty }\frac{H_{2k}}{k^2\:4^k}\binom{2k}{k}$$


First let's consider the following central binomial coefficient generating function. $$\sum _{k=1}^{\infty }\frac{x^{2k}}{k\:4^k}\binom{2k}{k}=-2\ln \left(1+\sqrt{1-x^2}\right)+2\ln \left(2\right)$$ $$-2\sum _{k=1}^{\infty }\frac{1}{k\:4^k}\binom{2k}{k}\int _0^1x^{2k-1}\ln \left(1-x\right)\:dx=4\int _0^1\frac{\ln \left(1-x\right)\ln \left(1+\sqrt{1-x^2}\right)}{x}\:dx$$ $$-4\ln \left(2\right)\int _0^1\frac{\ln \left(1-x\right)}{x}\:dx$$ $$\sum _{k=1}^{\infty }\frac{H_{2k}}{k^2\:4^k}\binom{2k}{k}=4\int _0^1\frac{\ln \left(1-x\right)\ln \left(1+\sqrt{1-x^2}\right)}{x}\:dx+4\ln \left(2\right)\zeta \left(2\right)$$


$$\int _0^1\frac{\ln \left(1-x\right)\ln \left(1+\sqrt{1-x^2}\right)}{x}\:dx=\int _0^1\frac{\operatorname{Li}_2\left(x\right)}{x}\:dx-\int _0^1\frac{\operatorname{Li}_2\left(x\right)}{x\sqrt{1-x^2}}\:dx$$ $$=\zeta \left(3\right)-\int _0^{\frac{\pi }{2}}\csc \left(x\right)\operatorname{Li}_2\left(\sin \left(x\right)\right)\:dx=\zeta \left(3\right)-\int _0^1\frac{\operatorname{Li}_2\left(\frac{2t}{1+t^2}\right)}{t}dt$$ $$=\zeta \left(3\right)+2\int _0^1\frac{t\ln \left(t\right)\ln \left(1-t\right)}{1+t^2}\:dt-\int _0^1\frac{t\ln \left(t\right)\ln \left(1+t^2\right)}{1+t^2}\:dt-2\int _0^1\frac{\ln \left(t\right)\ln \left(1-t\right)}{t}\:dt$$ $$+\int _0^1\frac{\ln \left(t\right)\ln \left(1+t^2\right)}{t}\:dt+2\int _0^1\frac{t\ln \left(t\right)\ln \left(1-t\right)}{1+t^2}\:dt-\int _0^1\frac{t\ln \left(t\right)\ln \left(1+t^2\right)}{1+t^2}\:dt$$ $$=\zeta \left(3\right)+4\underbrace{\int _0^1\frac{t\ln \left(t\right)\ln \left(1-t\right)}{1+t^2}\:dt}_{I}+\frac{1}{4}\underbrace{\int _0^1\frac{\ln ^2\left(1+t\right)}{t}\:dt}_{\frac{1}{4}\zeta \left(3\right)}$$ $$-2\underbrace{\int _0^1\frac{\ln \left(t\right)\ln \left(1-t\right)}{t}\:dt}_{\zeta \left(3\right)}+\underbrace{\int _0^1\frac{\ln \left(t\right)\ln \left(1+t^2\right)}{t}\:dt}_{-\frac{3}{16}\zeta \left(3\right)}$$

The integral $I$ can be found evaluated elegantly in the book (Almost) Impossible Integrals, Sums, and Series through pages $\#98,\#99,\#100$, making use of it's result we have: $$\int _0^1\frac{\ln \left(1-x\right)\ln \left(1+\sqrt{1-x^2}\right)}{x}\:dx=\frac{23}{16}\zeta \left(3\right)-\frac{9}{4}\ln \left(2\right)\zeta \left(2\right)$$ And in the process we also proved that: $$\int _0^1\frac{\operatorname{Li}_2\left(x\right)}{x\sqrt{1-x^2}}\:dx=-\frac{7}{16}\zeta \left(3\right)+\frac{9}{4}\ln \left(2\right)\zeta \left(2\right)$$


Collecting the results we finally have: $$\sum _{k=1}^{\infty }\frac{H_{2k}}{k^2\:4^k}\binom{2k}{k}=\frac{23}{4}\zeta \left(3\right)-5\ln \left(2\right)\zeta \left(2\right)$$

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