# Infinite Series $\sum_{n=1}^{\infty}\frac{4^nH_n}{n^2{2n\choose n}}$

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:

$$S=2\int_{0}^{1}\frac{x}{1-x^2}\left(\frac{\pi^2}{2}-2\arcsin^2(x)\right)dx$$

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 $$\forall x\in[0,1],\frac{1-x^n}{1-x}), we get: $$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 $$\displaystyle\sum\limits_{n=1}^{\infty}\frac{(4x)^n}{n^2{2n\choose n}}=2\arcsin^2\left(\sqrt{x}\right)$$

So,

$$S=\int_{0}^{1}\frac{1}{1-x}\left(\frac{\pi^2}{2}-2\arcsin^2\left(\sqrt{x}\right)\right)dx$$

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?

• 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. Commented Jul 28, 2019 at 18:53
• 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 $$\tilde{S} := \sum_{n = 1}^{\infty} \frac{H_n \cdot n!}{n^2 \cdot \Gamma\left(n + \frac{1}{2}\right)}.$$ Commented Jul 28, 2019 at 18:56
• 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. Commented Jul 28, 2019 at 19:09
• ㄴ ㄱ - I made a mistake earlier. I have now an integral representation, and I tested it numerically. Check my edit. Commented Jul 28, 2019 at 19:26
• ㄴ ㄱ - Thanks for your answer. As for the derivation of the integral representation, see my edit. Commented Jul 28, 2019 at 23:00

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

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