Euler Sums系列（一）

\[\Large\sum_{n=1}^{\infty} \frac{H_{n}}{2^nn^4}\]

\(\Large\mathbf{Solution:}\)
Let
\[\mathcal{S}=\sum^\infty_{n=1}\frac{H_n}{n^42^n}\]
We first consider a slightly different yet related sum. The main idea is to solve this sum with two different methods, one of which involves the sum in question. This then allows us to determine the value of the desired sum.
\[\begin{align*}
&\sum^\infty_{n=1}\frac{(-1)^nH_n}{n^4}
=\frac{1}{6}\sum^\infty_{n=1}(-1)^{n-1}H_n\int^1_0x^{n-1}\ln^3{x}\ {\rm d}x
=\frac{1}{6}\int^1_0\frac{\ln^3{x}\ln(1+x)}{x(1+x)}{\rm d}x\\
=&\frac{1}{6}\int^1_0\frac{\ln^3{x}\ln(1+x)}{x}{\rm d}x-\frac{1}{6}\int^1_0\frac{\ln^3{x}\ln(1+x)}{1+x}{\rm d}x
=\frac{1}{6}\sum^\infty_{n=1}\frac{(-1)^{n-1}}{n}\int^1_0x^{n-1}\ln^3{x}\ {\rm d}x\\
-&\frac{1}{6}\int^2_1\frac{\ln{x}\ln^3(x-1)}{x}{\rm d}x
=\sum^\infty_{n=1}\frac{(-1)^{n}}{n^5}+\int^1_{\frac{1}{2}}\frac{\ln{x}\ln^3(1-x)}{6x}-\int^1_{\frac{1}{2}}\frac{\ln^2{x}\ln^2(1-x)}{2x}{\rm d}x\\+&\int^1_{\frac{1}{2}}\frac{\ln^3{x}\ln(1-x)}{2x}{\rm d}x-\int^1_{\frac{1}{2}}\frac{\ln^4{x}}{6x}{\rm d}x
=-\frac{15}{16}\zeta(5)+\mathcal{I}_1-\mathcal{I}_2+\mathcal{I}_3-\mathcal{I}_4
\end{align*}\]
Starting with the easiest integral,
\[\begin{align*}
\mathcal{I}_4=\frac{1}{30}\ln^5{2}
\end{align*}\]
For \(\mathcal{I}_3\),
\[\begin{align*}
\mathcal{I}_3
=&-\frac{1}{2}\sum^\infty_{n=1}\frac{1}{n}\int^1_{\frac{1}{2}}x^{n-1}\ln^3{x}\ {\rm d}x
=-\frac{1}{2}\sum^\infty_{n=1}\frac{1}{n}\frac{\partial^3}{\partial n^3}\left(\frac{1}{n}-\frac{1}{n2^n}\right)\\
=&\sum^\infty_{n=1}\left(\frac{3}{n^5}-\frac{3}{n^52^n}-\frac{3\ln{2}}{n^42^n}-\frac{3\ln^2{2}}{n^32^{n+1}}-\frac{\ln^3{2}}{n^22^{n+1}}\right)\\
=&3\zeta(5)-3{\rm Li}_5\left(\dfrac{1}{2}\right)-3{\rm Li}_4\left(\dfrac{1}{2}\right)\ln{2}-\frac{3}{2}\ln^2{2}\left(\frac{7}{8}\zeta(3)-\frac{\pi^2}{12}\ln{2}+\frac{1}{6}\ln^3{2}\right)\\&-\frac{1}{2}\ln^3{2}\left(\frac{\pi^2}{12}-\frac{1}{2}\ln^2{2}\right)\\
=&3\zeta(5)-3{\rm Li}_5\left(\dfrac{1}{2}\right)-3{\rm Li}_4\left(\dfrac{1}{2}\right)\ln{2}-\frac{21}{16}\zeta(3)\ln^2{2}+\frac{\pi^2}{12}\ln^3{2}
\end{align*}\]
For \(\mathcal{I}_2\),
\[\begin{align*}
\mathcal{I}_2
=&\frac{1}{6}\ln^5{2}+\frac{1}{3}\int^1_{\frac{1}{2}}\frac{\ln^3{x}\ln(1-x)}{1-x}{\rm d}x
=\frac{1}{6}\ln^5{2}-\frac{1}{3}\sum^\infty_{n=1}H_n\frac{\partial^3}{\partial n^3}\left(\frac{1}{n+1}-\frac{1}{(n+1)2^{n+1}}\right)\\
=&\frac{1}{6}\ln^5{2}+\sum^\infty_{n=1}\frac{2H_n}{(n+1)^4}-\sum^\infty_{n=1}\frac{2H_n}{(n+1)^42^{n+1}}-\sum^\infty_{n=1}\frac{2\ln{2}H_n}{(n+1)^32^{n+1}}\\
&-\sum^\infty_{n=1}\frac{\ln^2{2}H_n}{(n+1)^22^{n+1}}-\sum^\infty_{n=1}\frac{\ln^3{2}H_n}{3(n+1)2^{n+1}}\\
=&\frac{1}{6}\ln^5{2}+4\zeta(5)-\frac{\pi^2}{3}\zeta(3)-2\mathcal{S}+2{\rm Li}_5\left(\dfrac{1}{2}\right)-\frac{\pi^4}{360}\ln{2}+\frac{1}{4}\zeta(3)\ln^2{2}-\frac{1}{12}\ln^5{2}\\
&-\frac{1}{8}\zeta(3)\ln^2{2}+\frac{1}{6}\ln^5{2}-\frac{1}{6}\ln^5{2}\\
=&-2\mathcal{S}+2{\rm Li}_5\left(\dfrac{1}{2}\right)+4\zeta(5)-\frac{\pi^4}{360}\ln{2}+\frac{1}{8}\zeta(3)\ln^2{2}-\frac{\pi^2}{3}\zeta(3)+\frac{1}{12}\ln^5{2}
\end{align*}\]
For \(\mathcal{I}_1\),
\[\begin{align*}
\mathcal{I}_1
=&\frac{1}{6}\int^{\frac{1}{2}}_0\frac{\ln^3{x}\ln(1-x)}{1-x}{\rm d}x
=-\frac{1}{6}\sum^\infty_{n=1}H_n\frac{\partial^3}{\partial n^3}\left(\frac{1}{(n+1)2^{n+1}}\right)\\
=&\sum^\infty_{n=1}\frac{H_n}{(n+1)^42^{n+1}}+\sum^\infty_{n=1}\frac{\ln{2}H_n}{(n+1)^32^{n+1}}+\sum^\infty_{n=1}\frac{\ln^2{2}H_n}{2(n+1)^22^{n+1}}+\sum^\infty_{n=1}\frac{\ln^3{2}H_n}{6(n+1)2^{n+1}}\\
=&\mathcal{S}-{\rm Li}_5\left(\dfrac{1}{2}\right)+\frac{\pi^4}{720}\ln{2}-\frac{1}{16}\zeta(3)\ln^2{2}+\frac{1}{24}\ln^5{2}
\end{align*}\]
Combining these four integrals as \(\mathcal{I}_1-\mathcal{I}_2+\mathcal{I}_3-\mathcal{I}_4\) and \(\displaystyle -\dfrac{15}{16}\zeta(5)\) gives
\[\begin{align*}
\sum^\infty_{n=1}\frac{(-1)^nH_n}{n^4}
=&3\mathcal{S}-6{\rm Li}_5\left(\dfrac{1}{2}\right)-\frac{31}{16}\zeta(5)-3{\rm Li}_4\left(\dfrac{1}{2}\right)\ln{2}+\frac{\pi^4}{240}\ln{2}\\&-\frac{3}{2}\zeta(3)\ln^2{2}+\frac{\pi^2}{3}\zeta(3)+\frac{\pi^2}{12}\ln^3{2}-\frac{3}{40}\ln^5{2}
\end{align*}\]
But consider \(\displaystyle f(z)=\frac{\pi\csc(\pi z)(\gamma+\psi_0(-z))}{z^4}\). At the positive integers,
\[\sum^\infty_{n=1}{\rm Res}(f,n)=\sum^\infty_{n=1}\operatorname*{Res}_{z=n}\left[\frac{(-1)^n}{z^4(z-n)^2}+\frac{(-1)^nH_n}{z^4(z-n)}\right]=\sum^\infty_{n=1}\frac{(-1)^nH_n}{n^4}+\frac{15}{4}\zeta(5)\]
At \(z=0\),
\[\begin{align*}
{\rm Res}(f,0)
&=[z^3]\left(\frac{1}{z}+\frac{\pi^2}{6}z+\frac{7\pi^4}{360}z^3\right)\left(\frac{1}{z}-\frac{\pi^2}{6}z-\zeta(3)z^2-\frac{\pi^4}{90}z^3-\zeta(5)z^4\right)\\
&=-\zeta(5)-\frac{\pi^2}{6}\zeta(3)
\end{align*}\]
At the negative integers,
\[\begin{align*}
\sum^\infty_{n=1}{\rm Res}(f,-n)
&=\sum^\infty_{n=1}\frac{(-1)^nH_n}{n^4}+\frac{15}{16}\zeta(5)
\end{align*}\]
Since the sum of the residues is zero,
\[\sum^\infty_{n=1}\frac{(-1)^nH_n}{n^4}=-\frac{59}{32}\zeta(5)+\frac{\pi^2}{12}\zeta(3)\]
Hence,
\[\begin{align*}
-\frac{59}{32}\zeta(5)+\frac{\pi^2}{12}\zeta(3)
=&3\mathcal{S}-6{\rm Li}_5\left(\dfrac{1}{2}\right)-\frac{31}{16}\zeta(5)-3{\rm Li}_4\left(\dfrac{1}{2}\right)\ln{2}+\frac{\pi^4}{240}\ln{2}\\&-\frac{3}{2}\zeta(3)\ln^2{2}+\frac{\pi^2}{3}\zeta(3)+\frac{\pi^2}{12}\ln^3{2}-\frac{3}{40}\ln^5{2}
\end{align*}\]
This implies that
\[\boxed{\begin{align*}
{\sum^\infty_{n=1}\frac{H_n}{n^42^n}}
{=}&\color{blue}{2{\rm Li}_5\left(\dfrac{1}{2}\right)+\frac{1}{32}\zeta(5)+{\rm Li}_4\left(\dfrac{1}{2}\right)\ln{2}-\frac{\pi^4}{720}\ln{2}+\frac{1}{2}\zeta(3)\ln^2{2}}\\&\color{blue}{-\frac{\pi^2}{12}\zeta(3)-\frac{\pi^2}{36}\ln^3{2}+\frac{1}{40}\ln^5{2}}
\end{align*}}\]

巴特西

Euler Sums系列（一）

最新文章

热门文章