sequences and series - Generalized Euler sum $sum_{n=1}^infty frac{H_n}{n^q}$

I found the following formula

$$\sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)$$

and it is cited that Euler proved the formula above , but how ?

Do there exist other proofs ?

Can we have a general formula for the alternating form

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

Answer

$$

\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

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

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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