# Euler Sum

Sums of the form $\displaystyle E(p,q)=\sum_{n=1}^\infty\frac{H_n^{(p)}}{n^q}$ where $\displaystyle H_n^{(p)}=\sum_{m=1}^n\frac{1}{m^p}$, sometimes are called Euler sums. There are several ways to evaluate these  sums. Aabout 240 years ago, Leonhard Euler(1707-1783) in 1775 proved that for $q\geq2$,

$\boxed{\displaystyle{\sum_{n=1}^\infty\frac{H_n}{n^q}=\frac{q+2}{2}\zeta(q+1)-\frac{1}{2}\sum_{r=1}^{q-2}\zeta(q-r)\zeta(r+1)}}$

In the following you can see an elementary proof of this formula in the three steps:

Step 1:

$\displaystyle\sum_{n=1}^\infty\frac{H_n}{n^q}=\sum_{n=1}^\infty\sum_{m=1}^n\frac{1}{mn^q}$

$\displaystyle=\sum_{m=1}^\infty\sum_{n=m}^\infty\frac{1}{mn^q}$

$\displaystyle=\zeta(q+1)+\sum_{m=1}^\infty\sum_{n=m+1}^\infty\frac{1}{mn^q}$

$\displaystyle=\zeta(q+1)+\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{1}{m(n+m)^q}$

$\displaystyle=\zeta(q+1)+\frac{1}{2}\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{1}{m(n+m)^q}+\frac{1}{n(n+m)^q}$

Hence,

$\boxed{\displaystyle{\sum_{n=1}^\infty\frac{H_n}{n^q}=\zeta(q+1)+\frac{1}{2}\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{1}{m(n+m)^q}+\frac{1}{n(n+m)^q}}}$

Step 2:

$\displaystyle H_n=\sum_{m=1}^n\frac{1}{m}=\sum_{i=0}^{n-1}\sum_{m=1}^\infty\frac{1}{m+i}-\frac{1}{m+i+1}$

$\displaystyle=\sum_{m=1}^\infty\sum_{i=0}^{n-1}\frac{1}{m+i}-\frac{1}{m+i+1}$

$\displaystyle=\sum_{m=1}^\infty\frac{1}{m}-\frac{1}{n+m}$

Now by recent formula that is also an alternate definition of the Harmonic numbers, we can write:

$\boxed{\displaystyle\sum_{n=1}^\infty\frac{H_n}{n^q}=\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{1}{mn^q}-\frac{1}{(n+m)n^q}}$

Step 3:

$\displaystyle\sum_{n=1}^\infty\frac{H_n}{n^q}$ $\displaystyle=2\sum_{n=1}^\infty\frac{H_n}{n^q}$ $\displaystyle-\sum_{n=1}^\infty\frac{H_n}{n^q}$

$\displaystyle{=2\zeta(q+1)+\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{1}{m(n+m)^q}+\frac{1}{n(n+m)^q}-\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{1}{mn^q}-\frac{1}{(n+m)n^q}}$

$\displaystyle=2\zeta(q+1)+\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{1}{m(n+m)^{q-1}n}-\frac{1}{(n+m)n^{q-1}m}$

$\displaystyle=2\zeta(q+1)+\sum_{n=1}^\infty\sum_{m=n+1}^\infty\frac{1}{nm^{q-1}(m-n)}-\frac{1}{mn^{q-1}(m-n)}$

$\displaystyle=2\zeta(q+1)-\frac{1}{2}\sum_{\substack{m,n=1\\m\ne n}}^\infty\frac{1}{mn^{q-1}(m-n)}-\frac{1}{nm^{q-1}(m-n)}$

$\displaystyle=2\zeta(q+1)-\frac{1}{2}\sum_{\substack{m,n=1\\m\ne n}}^\infty\frac{m^{q-2}-n^{q-2}}{m^{q-1}n^{q-1}(m-n)}$

$\displaystyle=2\zeta(q+1)-\frac{1}{2}\sum_{\substack{m,n=1\\m\ne n}}^\infty\sum_{r=1}^{q-2}\frac{1}{m^{q-r}n^{r+1}}$

$\displaystyle=2\zeta(q+1)-\frac{1}{2}\left(\sum_{m,n=1}^\infty\sum_{r=1}^{q-2}\frac{1}{m^{q-r}n^{r+1}}-\sum_{\substack{m,n=1\\m=n}}^\infty\sum_{r=1}^{q-2}\frac{1}{m^{q-r}n^{r+1}}\right)$

$\displaystyle=2\zeta(q+1)+\frac{q-2}{2}\zeta(q+1)-\frac{1}{2}\sum_{r=1}^{q-2}\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{1}{m^{q-r}n^{r+1}}$

$\displaystyle{=\frac{q+2}{2}\zeta(q+1)-\frac{1}{2}\sum_{r=1}^{q-2}\zeta(q-r)\zeta(r+1)}.$