Another great Euler sum
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
\[ -\frac{1}{t} + \frac{\sqrt{1-t^2}}{t} + \arcsin(t) = \sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}(2n-1)^2}t^{2n-1} \]
\[ \Longrightarrow \sqrt{1-t^2} + t\arcsin(t) -1 = \sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}(2n-1)^2}t^{2n} \]
