I want to find the closed form of:

$\displaystyle \tag*{} \sum \limits _{n=1}^{\infty}\frac{(-1)^n (H_{2n}-H_{n})}{n2^n \binom{2n}{n}}$

Where $H_{k}$ is $k^{\text{th}}$ harmonic number

I tried to expand the numerator (Harmonic numbers) in terms of integral, to get:

$\displaystyle \tag*{} \sum \limits_{n=1}^{\infty} \frac{(-1)^n}{n2^n\binom{2n}{n}} \int _{0}^{1} \frac{x^n - x^{2n}}{1-x} \ \mathrm dx$

And I found that with the help of series expansion of $\sin^{-1}(x)$ and subsituting $x = i \sqrt{x} /8 $ where $i^2=1$

$\displaystyle \tag*{} -2(\sinh^{-1} (\sqrt{x}/8))^2 = \sum \limits_{n=1}^{\infty} \frac{(-1)^nx^n}{n^22^n \binom{2n}{n}} $

But this has $n^2$ in the denominator, which makes it complicated. EDIT: we can eliminate $n^2$ by differentiating and multiplying by $x$ as mentioned in the comments. But now, how can we solve our sum since $H_{2n}-H_n$ is numerator?

And I have the general formula for generating sum:

$ \displaystyle \tag*{} \sum \limits _{n=1}^{\infty} \frac{x^n}{n^y \binom{2n}{n}}$

And this doesn't have $n$ in the denominator and also it has closed-form $\forall \ y \geq 2$

Maybe if there is a way of expressing the denominator in the form integral, the sum can be changed in evaluating the double integral. I think there are other easy ways (such as using Hypergeometric functions)? Any help would be appreciated.

EDIT 2: From the help of comments and a quora user, $\DeclareMathOperator{\arcsinh}{arcsinh}$

By @Bertrand87, we have:

$\displaystyle \tag{1} H_{k} - H_{2k} + \ln (2) = \int _{0}^{1} \frac{x^{2k}}{1+x} \ \mathrm dx$

To make use of this, we express our sum as follows:

$\displaystyle \tag*{} S = \sum \limits_{k=1}^{\infty} \frac{(-1)^k(H_{2k}-H_k - \ln2)}{k2^k \binom{2k}{k}} + \sum \limits_{k=1}^{\infty} \frac{(-1)^k(\ln2)}{k2^k \binom{2k}{k}}$

We know

$\displaystyle \tag*{} 2\arcsin^2(x) = \sum \limits_{k=1}^{\infty} \frac{(2x)^{2k}}{k^2 \binom{2k}{k}}$

We differentiate both sides w.r.t $x$ both sides,

$\displaystyle \tag*{} \frac{2 \arcsin(x)}{\sqrt{1-x^2}} = \sum \limits_{k=1}^{\infty} \frac{(2x)^{2k-1}}{k \binom{2k}{k}}$

Now, we multiply both sides by $(2x)$ and define $x:= ix/ \sqrt{8}$ to get:

$\displaystyle \tag{2} \frac{-2x \arcsinh (x/ \sqrt {8})}{\sqrt{8}\sqrt{1+x^2/8}} = \sum \limits_{k=1}^{\infty} \frac{(-1)^k x^{2k}}{k2^k \binom{2k}{k}}$

We now multiply both sides by $-1/(1+x)$ and integrate from $0$ to $1$ and arrive at:

$\displaystyle \tag*{} \frac{2}{\sqrt {8}}\int_{0}^{1}\frac{x\arcsinh(x/ \sqrt{8})}{\sqrt{1+x^2/8} (1+x)} \ \mathrm dx = \sum \limits_{k=1}^{\infty} \frac{(-1)^k(H_{2k}-H_k - \ln2)}{k2^k \binom{2k}{k}}$

Similarly, from $(2)$ if we let $x=1$ and multiply both sides by $\ln 2$, it yields:

$\displaystyle \tag*{} \frac{-2 \arcsinh (1/ \sqrt{8}) \ln 2}{ \sqrt{8} \sqrt{1 + 1/8}} =\sum \limits_{k=1}^{\infty} \frac{(-1)^k(\ln2)}{k2^k \binom{2k}{k}} \approx -0.1601$

Now, our only problem is to evaluate the integral:

$\displaystyle \tag*{} \boxed{\frac{2}{\sqrt {8}}\int_{0}^{1}\frac{x\arcsinh(x/ \sqrt{8})}{\sqrt{1+x^2/8} (1+x)} \ \mathrm dx} $ Can anyone help me with this integral?

  • $\begingroup$ What is the meaning of $(2)^n$ (versus $2^n$)? $\endgroup$ Apr 7, 2022 at 14:47
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    $\begingroup$ Try differentiating the series expansion that you found once and then multiply through by $x$ to get rid of the $n^2$. $\endgroup$ Apr 7, 2022 at 14:49
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    $\begingroup$ @4444 I don't know, I'm thinking about it. $\endgroup$ Apr 7, 2022 at 15:12
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    $\begingroup$ Maybe the following integral representation is useful: $$H_{n} - H_{2n} + \ln(2) = \int_{0}^{1} \frac{x^{2n}}{1+x} dx $$ $\endgroup$
    – Bertrand87
    Apr 7, 2022 at 23:21
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    $\begingroup$ @Bertrand87 thanks, i have edited my answer due to your hint. $\endgroup$
    – Dhanvin
    Apr 9, 2022 at 6:11

4 Answers 4


$\newcommand{arcsinh}{\operatorname{arcsinh}}\newcommand{csch}{\operatorname{csch}}\newcommand{arctanh}{\operatorname{arctanh}}\newcommand{Li}{\operatorname{Li}}$ Claim. $\displaystyle\int_0^1\frac{x\arcsinh(x/\sqrt8)}{(1+x)\sqrt{1+x^2/8}}\,dx=\frac{24\log^22-\pi^2}{18\sqrt2}$.

Proof: Substituting $u=\arcsinh(x/\sqrt8)$ gives \begin{align}\frac18\int_0^1\frac{x\arcsinh(x/\sqrt8)}{(1+x)\sqrt{1+x^2/8}}\,dx&=\int_0^{\log\sqrt2}\frac u{\sqrt8+\csch u}\,du.\end{align} Substituting $v=\tanh(u/2)$ and using partial fractions, we find \begin{align}\int\frac{du}{\sqrt8+\csch u}&=\int\frac{4v}{(v^2-1)(v^2-4\sqrt2v-1)}\,dv\\&=\frac1{\sqrt2}\int\left(\frac1{v^2-4\sqrt2v-1}-\frac1{v^2-1}\right)\,dv\\&=\frac{3u+2\arctanh([2\sqrt2-\tanh(u/2)]/3)}{6\sqrt2}+C\end{align} and from integration by parts, \begin{align}\int_0^{\log\sqrt2}\frac u{\sqrt8+\csch u}\,du&=\small\frac{\log(24+16\sqrt2)\log\sqrt2}{6\sqrt2}-\int_0^{\log\sqrt2}\frac{3u+2\arctanh([2\sqrt2-\tanh(u/2)]/3)}{6\sqrt2}\,du\end{align} Therefore, \begin{align}\int_0^1\frac{x\arcsinh(x/\sqrt8)}{(1+x)\sqrt{1+x^2/8}}\,dx&=\frac{\sqrt2\log(24+16\sqrt2)\log2}3-\frac{\log^22}{2\sqrt2}\\&\quad-\frac{4\sqrt2}3\int_0^{\log\sqrt2}\arctanh\left(\frac{2\sqrt2-\tanh(u/2)}3\right)\,du.\end{align} Now \begin{align}\small2\arctanh\left(\frac{2\sqrt2-\tanh(u/2)}3\right)&=\small\log\left(3+2\sqrt2-\tanh\frac u2\right)-\log\left(3-2\sqrt2+\tanh\frac u2\right)\\&=\small\log\left((1+\sqrt2)e^u+(2+\sqrt2)\right)-\log\left((2-\sqrt2)e^u+(1-\sqrt2)\right)\\&=\small\log\left(\frac{e^u}{\sqrt2}+1\right)-\log\left(\sqrt2e^u-1\right)+\log(4+3\sqrt2)\end{align} so we have, after combining constants, \begin{align}\small\int_0^1\frac{x\arcsinh(x/\sqrt8)}{(1+x)\sqrt{1+x^2/8}}\,dx&=\small\frac{7\log^22}{6\sqrt2}-\frac{2\sqrt2}3\int_0^{\log\sqrt2}\log\left(\frac{e^u}{\sqrt2}+1\right)\,du+\frac{2\sqrt2}3\int_0^{\log\sqrt2}\log\left(\sqrt2e^u-1\right)\,du\\&=\small\frac{7\log^22}{6\sqrt2}+\frac{2\sqrt2}3\left(\Li_2(-1)-\Li_2\left(-\frac1{\sqrt2}\right)+\Li_2(\sqrt2)-\Li_2(2)-i\pi\log\sqrt2\right).\end{align} Since \begin{align}\Li_2(-1)&=-\frac{\pi^2}{12}\\\Li_2(\sqrt2)-\Li_2\left(-\frac1{\sqrt2}\right)&=\Li_2(\sqrt2)+\Li_2\left(\frac1{\sqrt2}\right)-\frac12\Li_2\left(\frac12\right)\\&=-\frac{\pi^2}6-\frac{\log^2(-\sqrt2)}2-\frac12\left(\frac{\pi^2}{12}-\frac{\log^22}2\right)\\&=\frac{7\pi^2}{24}+\frac{\log^22}8-\frac{i\pi\log2}2\\\Li_2(2)&=\frac{\pi^2}4-i\pi\log2\end{align} we finally obtain \begin{align}\int_0^1\frac{x\arcsinh(x/\sqrt8)}{(1+x)\sqrt{1+x^2/8}}\,dx&=\frac{7\log^22}{6\sqrt2}+\frac{2\sqrt2}3\left(-\frac{\pi^2}{12}+\frac{7\pi^2}{24}+\frac{\log^22}8-\frac{\pi^2}4\right)\\&=\frac{24\log^22-\pi^2}{18\sqrt2}.\tag*{$\square$}\end{align}


$$\color{royalblue}{\sum \limits _{n=1}^{\infty}\frac{(-1)^{n-1} (H_{2n}-H_{n})}{n2^n \binom{2n}{n}} = \frac{\pi ^2}{36}-\frac{\log ^2(2)}{3}}$$

Proof sketch: Suffices to find $S=\sum \limits _{n=1}^{\infty}\frac{(-1)^{n-1} (H_{2n}-H_{n-1})}{n2^n \binom{2n}{n}}$. Note that $$\int_0^1 \frac{x^n(1-x)^n}{x} \log x dx = \frac{1}{n\binom{2n}{n}} (H_{n-1}-H_{2n})$$ so $$S = \int_0^1 \frac{(1-x)\log x}{1+\frac{1}{2}x(1-x)} dx$$ whose antiderivative can be expressed via dilogarithm $\text{Li}_2$, using special value of $\text{Li}_2(1/2)$ completes the proof.

As direct generalizations: $$\small\begin{aligned}\sum \limits _{n=1}^{\infty}\frac{(-1)^{n-1} (H_{2n}-H_{n-1})}{n^22^n \binom{2n}{n}} &= -\frac{\zeta (3)}{8}-\frac{1}{6} \log ^3(2)+\frac{1}{12} \pi ^2 \log (2) \\ \sum \limits _{n=1}^{\infty}\frac{(-1)^{n-1} (H_{2n}-H_{n-1})}{n^32^n \binom{2n}{n}} &= -2 \text{Li}_4\left(\frac{1}{2}\right)-\frac{13}{8} \zeta (3) \log (2)+\frac{19 \pi ^4}{720}-\frac{1}{24} \log ^4(2)+\frac{1}{24} \pi ^2 \log ^2(2) \end{aligned}$$

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    $\begingroup$ +1, just perfect, pisco. By any chance can you tell me how you figured out the integral representation of $\frac{1}{n \binom{2n}{n}}(H_{n-1}-H_{2n})$ $\endgroup$
    – Dhanvin
    Apr 9, 2022 at 15:46
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    $\begingroup$ @Dhanvin he did by using the definition of beta function $B(n,m)=\int_0^1 x^{n-1} (1-x)^{m-1}dx$ where he differentiated w.r.t $n$ once and w.r.t $m$ once them let $m\to n+1$. $\endgroup$ Jun 8, 2022 at 10:18
  • $\begingroup$ @AliShadhar thanks Shadhar! Glad to see you reply me. I'm greatly inspired by your books (I am a high school student and it's a very helpful). I'm just curious: if you were to publish a new edition, is this question eligible to be in the book? $\endgroup$
    – Dhanvin
    Jun 8, 2022 at 15:27
  • $\begingroup$ @Dhanvin thank you :) ... yes I am working on the second edition its focused on more generalizations and simpler proofs. If I come up with a different solution I will add it there. Nice problem though. $\endgroup$ Jun 8, 2022 at 16:03
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    $\begingroup$ @AliShadhar You might also be interested in some equalities at the end of arxiv.org/pdf/2201.01676.pdf, as well as references therein. Just like many problems of you and Prof Cornel, they all now have mechanical proofs. $\endgroup$
    – pisco
    Jun 8, 2022 at 16:14

Simplify the integral \begin{align} &\int_0^1\frac{x\sinh^{-1}\frac x{\sqrt8}}{(1+x)\sqrt{1+\frac{x^2}8}}\,dx\\ =& \>\sqrt2 \left(\sinh^{-1}\frac x{\sqrt8}\right)^2\bigg|_0^1 - \int_0^1\frac{\sinh^{-1}\frac x{\sqrt8}}{(1+x)\sqrt{1+\frac{x^2}8}}\,\overset{x=\sqrt2(y-\frac1y)}{dx}\\ =&\>\frac1{2\sqrt2}\ln^22-2\sqrt2 \int_1^{\sqrt2}\frac {\ln y}{\sqrt2 y^2+y-\sqrt2 }dy\tag1 \end{align} With $\int_0^1 \frac{\ln t}{1+t}dt=-\frac{\pi^2}{12}$ and $\int_0^1 \frac{\ln t}{1-t}dt=-\frac{\pi^2}{6}$ \begin{align}\int_0^{1/2} \frac{\ln t}{1-t}dt\overset{ibp}=& -\ln^22+\int_0^{1/2} \frac{\ln (1-t)}{t}\overset{t\to1-t}{dt}\\ =&\>\frac12\left(-\ln^22+\int_0^{1} \frac{\ln t}{1-t} dt\right) =-\frac12\ln^22 -\frac{\pi^2}{12} \end{align} \begin{align} & \int_1^{\sqrt2}\frac {\ln y}{\sqrt2 y^2+y-\sqrt2 }dy\\ =& \>\frac13\bigg(\int_1^{\sqrt2}\frac{\ln y}{y-\frac1{\sqrt2}}\overset{y=\frac1{\sqrt2t}}{dy}- \int_1^{\sqrt2} \frac{\ln y}{y+{\sqrt2}} \overset{y={\sqrt2t}}{dy}\bigg)\\ =&\>\frac13\bigg( -\frac38\ln^22-\int_0^1 \frac{\ln t}{1+t}dt+\int_0^{1/2}\frac{\ln t}{1-t}dt-\int_0^{1/\sqrt2} \frac{2t\ln t}{1-t^2}\overset{t^2\to t}{dt} \bigg)\\ =&\>-\frac18\ln^22+\frac{\pi^2}{36} +\frac16 \int_0^{1/2}\frac{\ln t}{1-t}dt = \frac{\pi^2}{72}-\frac5{24}\ln^22 \end{align} Plug into (1) to obtain $$\int_0^1\frac{x\sinh^{-1}\frac x{\sqrt8}}{(1+x)\sqrt{1+\frac{x^2}8}}\,dx=\frac{2\sqrt2}{3}\left(\ln^22-\frac{\pi^2}{24}\right)$$


After some rather lengthy yet not particularly difficult manipulations I have concluded that the more general series can be expressed as $$\sum_{n=1}^\infty \frac{(-1)^{n+1}(H_{2n}-H_{n})}{n2^n\binom{2n}{n}}x^{2n}=\frac{8x}{\sqrt{8+x^2}}\int_0^x\frac{\operatorname{arsinh}^2\frac{t}{2\sqrt2}}{t^2\sqrt{8+t^2}}dt,~~~\lvert x\rvert\leqslant2\sqrt2.$$ Hence, we may evaluate your given series by plugging in $x=1$, producing the result $$\sum_{n=1}^\infty \frac{(-1)^{n+1}(H_{2n}-H_{n})}{n2^n\binom{2n}{n}}=\frac{8}{3}\int_0^1\frac{\operatorname{arsinh}^2\frac{t}{2\sqrt2}}{t^2\sqrt{8+t^2}}dt.$$ This integral may be easier to work with than the integral that you arrived at, but at this moment in time I am unsure how to manipulate it myself. I have managed to show that $$\begin{align}\frac{8}{3}\int_0^1\frac{\operatorname{arsinh}^2\frac{t}{2\sqrt2}}{t^2\sqrt{8+t^2}}dt&=-\frac{1}{4}\log^22+\frac{2}{3}\int_0^1\frac{\operatorname{arsinh}\frac{t}{2\sqrt2}}{t}dt\\ &=-\frac{1}{4}\log^22-\frac{4\sqrt2}{3}\int_0^1\frac{\log t}{\sqrt{8+t^2}}dt\end{align}$$ using integration by parts but I'm not really sure where to go from there.

If you would like to know how to obtain these integrals then please post a comment. I'm afraid it's quite long so I won't show my full working in all likelihood.

I hope that you find this helpful. If you have any questions then please don't hesitate to ask.


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