Euler's work XIX

Another Euler sum involving the Catalan's constant $\beta(2)$

Today we show the proof of this nice Euler sum proposed by @Ali39342137 $$\sum_{n=1}^{\infty} \frac{\binom{2n}{n}\left(H_{2n}- H_{n} \right)}{2^{2n}(2n-1)^2} = \frac{3\pi\ln(2)}{2} + 2 - \pi - 2\beta(2)$$ For the proof we will use some properties of Harmonic numbers and Fourier series.

Proof

We need the following integral representation (proof in Appendix 1): $$H_{n} - H_{2n} + \ln(2) = \int_{0}^{1} \frac{x^{2n}}{1+x} dx \tag{1}$$ and the following series representation (proof in Appendix 2): $$\sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}(2n-1)^2}x^{2n} = \sqrt{1-x^2} + x\arcsin(x) -1 \quad |x|\leq 1 \tag{2}$$ Therefore $$\begin{align*} \sum_{n=1}^{\infty} \frac{\binom{2n}{n}\left(H_{2n}- H_{n} \right)}{2^{2n}(2n-1)^2} =& \sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}(2n-1)^2} \left( \ln(2) - \int_{0}^{1} \frac{x^{2n}}{1+x} dx \right) \quad \textrm{ from (1)} \\ =& \ln(2)\sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}(2n-1)^2}- \sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}(2n-1)^2}\int_{0}^{1} \frac{x^{2n}}{1+x} dx \end{align*}$$ From (2) if we put $x=1$ $$\sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}(2n-1)^2} = \arcsin(1) -1 = \frac{\pi}{2} - 1$$ Hence $$\begin{align*} \sum_{n=1}^{\infty} \frac{\binom{2n}{n}\left(H_{2n}- H_{n} \right)}{2^{2n}(2n-1)^2} =& \frac{\pi\ln(2)}{2} - \ln(2)-\int_{0}^{1} \frac{1}{1+x}\left(\sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}(2n-1)^2}x^{2n}\right) dx \\ =& \frac{\pi\ln(2)}{2} - \ln(2)- \int_{0}^{1} \frac{\sqrt{1-x^2} + x\arcsin(x) -1}{1+x} dx \quad \textrm{ from (2) }\\ =& \frac{\pi\ln(2)}{2} - \ln(2)- \underbrace{\int_{0}^{1} \frac{\sqrt{1-x^2}}{1+x} dx}_{I} - \underbrace{\int_{0}^{1}\frac{x\arcsin(x)}{1+x}dx}_{J} +\underbrace{\int_{0}^{1}\frac {1}{1+x} dx}_{K} \end{align*}$$ $$\begin{align*} I = \int_{0}^{1} \frac{\sqrt{1-x^2}}{1+x} dx =& 4\int_{0}^{1} \frac{w^2}{(1+w^2)^2} dw \quad \left( w \mapsto \frac{\sqrt{1-x^2}}{1+x} \right) \\ =& 4\int_{1}^{\infty} \frac{1}{(1+s^2)^2} ds \quad \left( s\mapsto \frac{1}{w} \right)\\ =& 4\int_{\frac{\pi}{2} }^{\frac{\pi}{2}} \frac{\sec^2(w)}{(1+\tan^2{w})^2} dw \quad \left(w \mapsto \arctan(s) \right) \\ =& 4\int_{\frac{\pi}{2} }^{\frac{\pi}{2}} \cos^2(w) dw \quad \left( \tan^2(\theta) +1 = \sec^2(\theta)\right) \\ =& 2\int_{\frac{\pi}{2} }^{\frac{\pi}{2}} \cos(2w)dw + \frac{1}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} dw \quad \left( 2\cos^2(\theta)-1 = \cos(2\theta) \right)\\ =& 2\int_{\frac{\pi}{4} }^{\frac{\pi}{2}} \cos(2w)dw + \frac{\pi}{8} \\ =& \int_{\frac{\pi}{2}}^{\pi} \cos(r)dr + \frac{\pi}{2} \quad (r \mapsto 2w)\\ =& -1 + \frac{\pi}{2}\\ \end{align*}$$ $$\begin{align*} J = \int_{0}^{1}\frac{x\arcsin(x)}{1+x}dx \stackrel{IBP}{=} & \left[x\arctan(x)-\ln(1+x)\arctan(x)\right]_{0}^{1} -\int_{0}^{1}\frac{x}{\sqrt{1-x^2}}dx +\int_{0}^{1} \frac{\ln(x+1)}{\sqrt{1-x^2}} dx \\ =& \frac{\pi}{2} -\frac{\pi\ln(2) }{2} -\left[-\sqrt{1-x^2}\right]_{0}^{1} +\int_{0}^{1} \frac{\ln(x+1)}{\sqrt{1-x^2}}dx\\ =& \frac{\pi}{2} -\frac{\pi\ln(2) }{2} -1 +\int_{0}^{1} \frac{\ln(x+1)}{\sqrt{1-x^2}}dx\\ =& \frac{\pi}{2} -\frac{\pi\ln(2) }{2} -1 +\int_{0}^{\frac{\pi}{2}} \ln(1+\cos(w))dw \quad (w \mapsto \operatorname{arccos}(x))\\ =& \frac{\pi}{2} -\frac{\pi\ln(2) }{2} -1 +\int_{0}^{\frac{\pi}{2}} \ln\left(2\cos^2\left(\frac{w}{2}\right)\right)dw \quad \left(\cos(2\theta) = 2\cos^2(\theta)-1 \right)\\ =& \frac{\pi}{2} -\frac{\pi\ln(2) }{2} -1 +\ln(2) \int_{0}^{\frac{\pi}{2}} dw +2\int_{0}^{\frac{\pi}{2}} \ln\left(\cos\left(\frac{w}{2}\right)\right)dw \\ =& \frac{\pi}{2} -1 -2\int_{0}^{\frac{\pi}{2}} \ln\left(\cos\left(\frac{w}{2}\right)\right)dw \end{align*}$$ Recall the Fourier series: $$\sum_{k=1}^{\infty} \frac{(-1)^{k-1}\cos(kx)}{k} = \ln\left(2\cos\left(\frac{x}{2}\right)\right)$$ Therefore $$\begin{align*} J=& \frac{\pi}{2} -1 -2\int_{0}^{\frac{\pi}{2}} \ln\left(\cos\left(\frac{w}{2}\right)\right)dw \\ =& \frac{\pi}{2} -1 -2\int_{0}^{\frac{\pi}{2}} \sum_{k=1}^{\infty} \frac{(-1)^{k-1}\cos(kw)}{k}dw -2\ln(2)\int_{0}^{\frac{\pi}{2}} dw\\ =& \frac{\pi}{2} -\pi\ln(2) -1 -2\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k}\int_{0}^{\frac{\pi}{2}} \cos(kw)dw \\ =& \frac{\pi}{2}-\pi\ln(2) -1 -2\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k^2}\int_{0}^{\frac{k\pi}{2}} \cos(r)dr \\ =& \frac{\pi}{2}-\pi\ln(2) -1 -2\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k^2}\left[\sin(r)\right]_{0}^{\frac{k\pi}{2}} \\ =& \frac{\pi}{2} -\pi\ln(2) -1 -2\sum_{k=1}^{\infty} \frac{(-1)^{k-1} \sin\left(\frac{k \pi}{2} \right)}{k^2} \\ \end{align*}$$ Note that $$\left(\sin\left(\frac{k \pi}{2}\right)\right)_{k\in \mathbb{N}} = (1,0,-1,0,1,0,-1,0,..)$$ Hence $$J = \frac{\pi}{2} -\pi\ln(2) -1 + 2\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{(2k-1)^2} = \frac{\pi}{2} -\pi\ln(2) -1 + 2\beta(2)$$ $$K = \int_{0}^{1} \frac{1}{1+x} dx = \left[\ln(1+x)\right]_{0}^{1} = \ln(2)$$ Therefore $$\begin{align*} \sum_{n=1}^{\infty} \frac{\binom{2n}{n}\left(H_{2n}- H_{n} \right)}{2^{2n}(2n-1)^2} =& \frac{\pi\ln(2)}{2} - \ln(2)- I - J + K \\ =& \frac{3\pi\ln(2)}{2} + 2 - \pi - 2\beta(2) \end{align*}$$ Hence, we can conclude $$\boxed{\sum_{n=1}^{\infty} \frac{\binom{2n}{n}\left(H_{2n}- H_{n} \right)}{2^{2n}(2n-1)^2} = \frac{3\pi\ln(2)}{2} + 2 - \pi - 2\beta(2) }$$

Appendix 1

$$\ln(2) + H_{n} - H_{2n} = \int_{0}^{1} \frac{x^{2n}}{1+x} dx$$

Proof

$$\begin{align*} \int_{0}^{1} \frac{x^{2n}}{1+x} dx\stackrel{IBP}{=}& \left[\ln(1+x)x^{2n}\right]_{0}^{1} - 2n\int_{0}^{1} \ln(1+x)x^{2n-1} dx \\ =& \ln(2) - 2n\int_{0}^{1}\sum_{j=1}^{\infty} \frac{(-1)^{j+1}}{j} x^{j} x^{2n-1} dx\\ =& \ln(2) - 2n\sum_{j=1}^{\infty} \frac{(-1)^{j+1}}{j} \int_{0}^{1} x^{j+2n-1} dx\\ =& \ln(2) + 2n\sum_{j=1}^{\infty} \frac{(-1)^{j}}{j(j+2n)} \\ =& \ln(2) + n\sum_{j=1}^{\infty} \frac{1}{j(j+n)} - 2n\sum_{j=1}^{\infty} \frac{1}{j(j+2n)}\\ \end{align*}$$ From the series expansion for Harmonic numbers: $$H_{x} = x \sum_{k=1}^{\infty} \frac{1}{k(x+k)}$$ we obtain $$\ln(2) + H_{n} - H_{2n} = \int_{0}^{1} \frac{x^{2n}}{1+x} dx$$

Appendix 2

$$\sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}(2n-1)^2}x^{2n} = \sqrt{1-x^2} + x\arcsin(x) -1 \quad |x|\leq 1$$

Proof

Recall the following basic series expansion: $$\frac{1}{\sqrt{1-x^2}} = \sum_{n=0}^{\infty} \frac{\binom{2n}{n}}{2^{2n}} x^{2n} \quad |x|\lt 1$$ Dividing by $x^2$ $$\frac{1}{x^2\sqrt{1-x^2}} = \sum_{n=0}^{\infty} \frac{\binom{2n}{n}}{2^{2n}} x^{2n-2} = \frac{1}{x^2} + \sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}} x^{2n-2}$$ $$\Longrightarrow \frac{1}{x^2\sqrt{1-x^2}}- \frac{1}{x^2} = \sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}} x^{2n-2}$$ Integrating from $0$ to $r$: $$\int_{0}^{r} \left[\frac{1}{x^2\sqrt{1-x^2}}- \frac{1}{x^2} \right] dx = \sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}} \int_{0}^{r}x^{2n-2}dx = \sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}(2n-1)}r^{2n-1}$$ - Note $$\begin{align*} \int \left[\frac{1}{x^2\sqrt{1-x^2}}- \frac{1}{x^2} \right] dx =& \int \frac{1}{x^2\sqrt{1-x^2}}dx- \int \frac{1}{x^2} dx\\ =& \int \frac{1}{x^2\sqrt{1-x^2}}dx+ \frac{1}{x} + C\\ =& \int \csc^2(w) dw + \frac{1}{x} + C \quad \left(x \mapsto \sin(w)\right)\\ =& -\cot(w) + \frac{1}{x} + C\\ =& -\cot(\arcsin(x)) + \frac{1}{x} + C\\ =& -\frac{\sqrt{1-x^2}}{x} + \frac{1}{x} + C \quad \left( \arcsin(x) = \operatorname{arccot}\left(\frac{\sqrt{1-x^2}}{x}\right)\right)\\ =& \frac{1-\sqrt{1-x^2}}{x} + C \end{align*}$$ Hence $$\begin{align*} \int_{0}^{r} \left[\frac{1}{x^2\sqrt{1-x^2}}- \frac{1}{x^2} \right] dx = & \left[ \frac{1-\sqrt{1-x^2}}{x}\right]_{0}^{r} \\ =& \frac{1-\sqrt{1-r^2}}{r} - \lim_{x \to 0 } \frac{1-\sqrt{1-x^2}}{x} \\ =& \frac{1-\sqrt{1-r^2}}{r} \end{align*}$$ Therefore $$\frac{1-\sqrt{1-r^2}}{r} = \sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}(2n-1)}r^{2n-1}$$ Dividing by $r$: $$\frac{1-\sqrt{1-r^2}}{r^2} = \sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}(2n-1)}r^{2n-2}$$ Integrating from $0$ to $t$: $$\int_{0}^{t} \frac{1-\sqrt{1-r^2}}{r^2} dt = \sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}(2n-1)}\int_{0}^{t} r^{2n-2} dt = \sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}(2n-1)^2}t^{2n-1}$$ Note that $$\begin{align*} \int \frac{1-\sqrt{1-r^2}}{r^2} dr =& \int \frac{1}{r^2} dr - \int \frac{\sqrt{1-r^2}}{r^2} dr \\ =& -\frac{1}{r} - \int \frac{\sqrt{1-r^2}}{r^2} dr + C\\ =& -\frac{1}{r} - \int \cot^2(w) dw + C \quad \left( r \mapsto \sin(w)\right)\\ =& -\frac{1}{r} - \int (\csc^2(w)-1) dw + C \quad \left( \csc^2(\theta) - \cot^2(\theta) = 1 \right)\\ =& -\frac{1}{r} + \cot(w) + w + C\\ =& -\frac{1}{r} + \cot(\arcsin(r)) + \arcsin(r) + C\\ =& -\frac{1}{r} + \frac{\sqrt{1-r^2}}{r} + \arcsin(r) + C \quad \left( \arcsin(x) = \operatorname{arccot}\left(\frac{\sqrt{1-x^2}}{x}\right)\right)\\ \end{align*}$$ Therefore $$\int_{0}^{t} \frac{1-\sqrt{1-r^2}}{r^2} dt = \left[-\frac{1}{r} + \frac{\sqrt{1-r^2}}{r} + \arcsin(r)\right]_{0}^{t} = -\frac{1}{t} - \frac{\sqrt{1-t^2}}{t} + \arcsin(t)$$ Hence $$-\frac{1}{t} + \frac{\sqrt{1-t^2}}{t} + \arcsin(t) = \sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}(2n-1)^2}t^{2n-1}$$ $$\Longrightarrow \sqrt{1-t^2} + t\arcsin(t) -1 = \sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}(2n-1)^2}t^{2n}$$