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)}.

Euler Sum