Euler Sum的若干研究

\begin{align}&\sum_{j=0}^k\zeta(k+2-j)\zeta(j+2)\\&=\sum_{m=1}^\infty\sum_{n=1}^\infty\sum_{j=0}^k\frac1{m^{k+2-j}n^{j+2}}\tag{1}\\&=(k+1)\zeta(k+4)+\sum_{\substack{m,n=1\\m\ne n}}^\infty\frac1{m^2n^2}\frac{\frac1{m^{k+1}}-\frac1{n^{k+1}}}{\frac1m-\frac1n}\tag{2}\\&=(k+1)\zeta(k+4)+\sum_{\substack{m,n=1\\m\ne n}}^\infty\frac1{nm^{k+2}(n-m)}-\frac1{mn^{k+2}(n-m)}\tag{3}\\&=(k+1)\zeta(k+4)+2\sum_{m=1}^\infty\sum_{n=m+1}^\infty\frac1{nm^{k+2}(n-m)}-\frac1{mn^{k+2}(n-m)}\tag{4}\\&=(k+1)\zeta(k+4)+2\sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{(n+m)m^{k+2}n}-\frac1{m(n+m)^{k+2}n}\tag{5}\\&=(k+1)\zeta(k+4)\\&+2\sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{m^{k+3}n}-\frac1{(m+n)m^{k+3}}\\&-2\sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{m(n+m)^{k+3}}+\frac1{n(n+m)^{k+3}}\tag{6}\\&=(k+1)\zeta(k+4)+2\sum_{m=1}^\infty\frac{H_m}{m^{k+3}}-4\sum_{n=1}^\infty\sum_{m=1}^\infty\frac1{n(n+m)^{k+3}}\tag{7}\\&=(k+1)\zeta(k+4)+2\sum_{m=1}^\infty\frac{H_m}{m^{k+3}}-4\sum_{n=1}^\infty\sum_{m=n+1}^\infty\frac1{nm^{k+3}}\tag{8}\\&=(k+1)\zeta(k+4)+2\sum_{m=1}^\infty\frac{H_m}{m^{k+3}}-4\sum_{n=1}^\infty\sum_{m=n}^\infty\frac1{nm^{k+3}}+4\zeta(k+4)\tag{9}\\&=(k+5)\zeta(k+4)+2\sum_{m=1}^\infty\frac{H_m}{m^{k+3}}-4\sum_{m=1}^\infty\sum_{n=1}^m\frac1{nm^{k+3}}\tag{10}\\&=(k+5)\zeta(k+4)+2\sum_{m=1}^\infty\frac{H_m}{m^{k+3}}-4\sum_{m=1}^\infty\frac{H_m}{m^{k+3}}\tag{11}\\&=(k+5)\zeta(k+4)-2\sum_{m=1}^\infty\frac{H_m}{m^{k+3}}\tag{12}\end{align}

 

Letting $q=k+3$ and reindexing $j\mapsto j-1$ yields

$$\sum_{j=1}^{q-2}\zeta(q-j)\zeta(j+1)=(q+2)\zeta(q+1)-2\sum_{m=1}^\infty\frac{H_m}{m^q}\tag{13}$$

and finally

$$\sum_{m=1}^\infty\frac{H_m}{m^q}=\frac{q+2}{2}\zeta(q+1)-\frac12\sum_{j=1}^{q-2}\zeta(q-j)\zeta(j+1)\tag{14}$$

 

Explanation

 

$\hphantom{0}(1)$ expand $\zeta$  

$\hphantom{0}(2)$ pull out the terms for $m=n$ and use the formula for finite geometric sums on the rest  

$\hphantom{0}(3)$ simplify terms  

$\hphantom{0}(4)$ utilize the symmetry of $\frac1{nm^{k+2}(n-m)}+\frac1{mn^{k+2}(m-n)}$  

$\hphantom{0}(5)$ $n\mapsto n+m$ and change the order of summation  

$\hphantom{0}(6)$ $\frac1{mn}=\frac1{m(m+n)}+\frac1{n(m+n)}$  

$\hphantom{0}(7)$ $H_m=\sum_{n=1}^\infty\frac1n-\frac1{n+m}$ and use the symmetry of $\frac1{m(n+m)^{k+3}}+\frac1{n(n+m)^{k+3}}$  

$\hphantom{0}(8)$ $m\mapsto m-n$  

$\hphantom{0}(9)$ subtract and add the terms for $m=n$  

$(10)$ combine $\zeta(k+4)$ and change the order of summation  

$(11)$ $H_m=\sum_{n=1}^m\frac1n$  

$(12)$ combine sums