I am trying to find a closed form for this infinite series: $$ S=\sum_{n=1}^{\infty}\frac{4^nH_n}{n^2{2n\choose n}}$$ Whith $H_n=\sum\limits_{k=1}^{n}\frac{1}{k}$ the harmonic numbers.

I found this integral representation of S:


Sketch of a proof:

Recall the integral representation of the harmonic numbers: $H_n=\displaystyle\int_{0}^{1}\frac{1-x^n}{1-x}dx$

By plugging it into the definition of S and interchanging the order of summation between $\displaystyle\sum$ and $\displaystyle\int$ (justified by the uniform convergence of the function series $\displaystyle\sum\left(x\to\frac{4^n}{n^2{2n\choose n}}\frac{1-x^n}{1-x}\right)$, because $\forall x\in[0,1],\frac{1-x^n}{1-x}<n$), we get: $$S=\int_{0}^{1}\frac{1}{1-x}\sum\limits_{n=1}^{\infty}\frac{4^n(1-x^n)}{n^2{2n\choose n}}dx$$ $$=\int_{0}^{1}\frac{1}{1-x}\left(\sum\limits_{n=1}^{\infty}\frac{4^n}{n^2{2n\choose n}}-\sum\limits_{n=1}^{\infty}\frac{(4x)^n}{n^2{2n\choose n}}\right)dx$$ $$=\int_{0}^{1}\frac{1}{1-x}\left(\frac{\pi^2}{2}-\sum\limits_{n=1}^{\infty}\frac{(4x)^n}{n^2{2n\choose n}}\right)dx$$ Using the result $\displaystyle\sum\limits_{n=1}^{\infty}\frac{4^n}{n^2{2n\choose n}}=\frac{\pi^2}{2}$.

At that point, we will rely on the taylor series expansion of $\arcsin^2$: $$\arcsin^2(x)=\frac{1}{2}\sum\limits_{n=1}^{\infty}\frac{4^n}{n^2{2n\choose n}}x^{2n}, |x|<1$$ Out of which we get $\displaystyle\sum\limits_{n=1}^{\infty}\frac{(4x)^n}{n^2{2n\choose n}}=2\arcsin^2\left(\sqrt{x}\right)$



Which, through the substitution $u=\sqrt{x}$, gives the integral representation above.

But beyond that, nothing so far. I tried to use the integral representation of $\frac{H_n}{n}$ to switch the order of summation, but it didn't lead anywhere. Any suggestion?

  • 1
    $\begingroup$ ㄴ ㄱ - Lately I'm enjoying studying infinite series that involve harmonic numbers and/or central binomial coefficients ${2n\choose n}$. I've already found closed forms for a fair amount of them, but this one is quite hard, due to the central binomial coefficient being on the denominator. Other than that, there is actually not much more context. $\endgroup$ Commented Jul 28, 2019 at 18:53
  • $\begingroup$ One idea is to notice $$\binom{2n}{n} = \frac{4^n \Gamma(n + 0.5)}{\sqrt{\pi} \cdot \Gamma(n + 1)},$$ so that cancels with the $4^n$ in the denominator. Dropping constants, you can now look at $$\tilde{S} := \sum_{n = 1}^{\infty} \frac{H_n \cdot n!}{n^2 \cdot \Gamma\left(n + \frac{1}{2}\right)}.$$ $\endgroup$ Commented Jul 28, 2019 at 18:56
  • $\begingroup$ ㄴ ㄱ - You're right, the integral diverges so it turns out my integral representation is simply wrong. I interchanged the order of summation between the series and the integral representation of $H_n$ without paying attention to the justification of that interchanging, which in this case is probably impossible. $\endgroup$ Commented Jul 28, 2019 at 19:09
  • 1
    $\begingroup$ ㄴ ㄱ - I made a mistake earlier. I have now an integral representation, and I tested it numerically. Check my edit. $\endgroup$ Commented Jul 28, 2019 at 19:26
  • 1
    $\begingroup$ ㄴ ㄱ - Thanks for your answer. As for the derivation of the integral representation, see my edit. $\endgroup$ Commented Jul 28, 2019 at 23:00

2 Answers 2


$$S=2\int_{0}^{1}\frac{x}{1-x^2}\left(\frac{\pi^2}{2}-2\arcsin^2(x)\right)dx\overset{IBP}=-4\int_0^1 \frac{\arcsin x\ln(1-x^2)}{\sqrt{1-x^2}}dx$$ $$\overset{x=\sin t}=-8\int_0^\frac{\pi}{2} t \ln(\cos t)dt=8 \ln 2 \int_0^\frac{\pi}{2}t dt+8\sum_{n=1}^\infty \frac{(-1)^n}{n}\int_0^\frac{\pi}{2} t\cos(2n t)dt$$ $$={\pi^2}\ln 2+2\sum_{n=1}^\infty \frac{1-(-1)^n}{n^3}=\boxed{\pi^2 \ln 2 +\frac72 \zeta(3)}$$

  • 1
    $\begingroup$ Nice answer, and great closed form ! (+1) $\endgroup$ Commented Jul 28, 2019 at 22:32
  • $\begingroup$ Could you please explain the series expansion for $\ln(\cos(t))$? $\endgroup$ Commented Jul 29, 2019 at 11:50
  • $\begingroup$ I linked the series expansion there (look here math.stackexchange.com/a/292477/515527 otherwise). $\endgroup$
    – Zacky
    Commented Jul 29, 2019 at 12:20
  • 2
    $\begingroup$ Thanks! Embarrassingly, I didn't notice that the blue formula was a hyperlink. $\endgroup$ Commented Jul 29, 2019 at 12:36

From here, we have

$$\frac{\arcsin z}{\sqrt{1-z^2}}=\sum_{n=1}^\infty\frac{(2z)^{2n-1}}{n{2n \choose n}}$$

substitute $z=\sqrt{y}$, we get

$$\sum_{n=1}^\infty\frac{4^ny^n}{n{2n \choose n}}=2\sqrt{y}\frac{\arcsin\sqrt{y}}{\sqrt{1-y}}$$

Now multiply both sides by $-\frac{\ln(1-y)}{y}$ then integrate from $y=0$ to $1$ and using the fact that $-\int_0^1 y^{n-1}\ln(1-x)\ dy=\frac{H_n}{n}$, we get

\begin{align} \sum_{n=1}^\infty\frac{4^nH_n}{n^2{2n \choose 2}}&=-2\int_0^1\frac{\arcsin\sqrt{y}}{\sqrt{y}\sqrt{1-y}}\ln(1-y)\ dy\overset{\arcsin\sqrt{y}=x}{=}-8\int_0^{\pi/2}x\ln(\cos x)\ dx\\ &=-8\int_0^{\pi/2}x\left\{-\ln2-\sum_{n=1}^\infty\frac{(-1)^n\cos(2nx)}{n}\right\}\ dx\\ &=\pi^2\ln2+8\sum_{n=1}^\infty\frac{(-1)^n}{n}\int_0^{\pi/2}x\cos(2nx) dx\\ &=\pi^2\ln2+8\sum_{n=1}^\infty\frac{(-1)^n}{n}\left(\frac{\pi\sin(n\pi)}{4n}+\frac{\cos(n\pi)}{4n^2}-\frac1{4n^2}\right)\\ &=\pi^2\ln2+2\pi\sum_{n=1}^\infty\frac{(-1)^n\sin(n\pi)}{n^2}+2\sum_{n=1}^\infty\frac{(-1)^n\cos(n\pi)}{n^3}-2\sum_{n=1}^\infty\frac{(-1)^n}{n^3}\\ &=\pi^2\ln2+0+2\sum_{n=1}^\infty\frac{(-1)^n(-1)^n}{n^3}-2\operatorname{Li}_3(-1)\\ &=\pi^2\ln2+2\zeta(3)-2\left(-\frac34\zeta(3)\right)\\ &=\pi^2\ln2+\frac72\zeta(3) \end{align}


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