4
$\begingroup$

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 ?

$\endgroup$
  • $\begingroup$ Don't you mean $H_n=\sum_{k=1}^{n}\frac{1}{k}$? $\endgroup$ – James Arathoon Jan 24 at 1:13
  • $\begingroup$ Oh yes sorry, now it's corrected ! $\endgroup$ – Harmonic Sun Jan 24 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? $\endgroup$ – Zachary Jan 24 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... $\endgroup$ – Harmonic Sun Jan 24 at 2:32
6
$\begingroup$

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 $a \in [0,1)$ .

$\endgroup$
  • 1
    $\begingroup$ This is a nice answer. Now, when I look at my junk, I am just ashamed. $\endgroup$ – Claude Leibovici Jan 24 at 5:44
  • 1
    $\begingroup$ Yes indeed, this is a very nice answer. $\endgroup$ – omegadot Jan 24 at 5:46
  • $\begingroup$ Thank you very much, this is a great answer ! $\endgroup$ – Harmonic Sun Jan 24 at 15:03
1
$\begingroup$

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 $$S_\infty=S_p+\sum_{n=p+1}^\infty b_n$$ $$\left( \begin{array}{cc} p & S_\infty \approx \\ 100 & 1.2109213325 \\ 200 & 1.2109203863 \\ 300 & 1.2109202368 \\ 400 & 1.2109201900 \\ 500 & 1.2109201700 \\ 600 & 1.2109201590 \\ 700 & 1.2109201535 \\ 800 & 1.2109201498 \\ 900 & 1.2109201475 \\ 1000 & 1.2109201458 \end{array} \right)$$

$\endgroup$
  • $\begingroup$ 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. $$\text{FullSimplify}\left[2 \sum _{m=1}^5 \left(H_{2 m}-H_{2 m-2}\right) \int_0^1 \text{FullSimplify}\left[\sum _{n=m}^{\infty } \frac{\binom{2 n}{n} x^{2 n-1}}{n 2^{2 n}}\right] \, dx\right]=\frac{11 \left(515328 \pi ^2-859301-6183936 \log ^2(2)\right)}{11612160}$$ $\endgroup$ – James Arathoon Jan 24 at 12:39
  • $\begingroup$ or for slightly simpler integral $$\text{FullSimplify}\left[2 \sum _{m=1}^5 \left(H_{2 m}-H_{2 m-2}\right) \int_0^{\frac{1}{2}} \text{FullSimplify}\left[\sum _{n=m}^{\infty } \frac{\binom{2 n}{n} x^{2 n-1}}{n}\right] \, dx\right] $$ $\endgroup$ – James Arathoon Jan 24 at 13:20
  • $\begingroup$ First comment should read after partial term 10 $\endgroup$ – James Arathoon Jan 24 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 :-) $\endgroup$ – Claude Leibovici Jan 24 at 16:00
1
$\begingroup$

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 $\ \displaystyle\int_0^1\frac{\operatorname{Li}_2(x)}{x\sqrt{1-x^2}}\ dx=\frac{3}{8}\pi^2 \ln(2) - \frac{7}{16} \zeta(3)$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.