Let's define
$$\sigma(m,n)=\sum_{k=1}^\infty\frac{H_k^{(m)}}{k^n}$$
where $H_k^{(m)}=\sum_{n=1}^{k}\frac{1}{n^m}$ is the k-th generalized harmonic number of order $m$.
In mathworld site Eq (20), I found
$$\sigma(m\text{ even},n\text{ odd})=\frac12\left[\binom{m+n}{m}+1\right]\zeta(m+n)+\zeta(m)\zeta(n)$$
$$-\sum_{j=1}^{m+n}\left[\binom{2j-2}{m-1}+\binom{2j-2}{n-1}\right]\zeta(2j-1)\zeta(m+n-2j+1)\label{1}\tag{1}$$
and
$$\sigma(m\text{ odd},n\text{ even})=-\frac12\left[\binom{m+n}{m}+1\right]\zeta(m+n)$$
$$+\sum_{j=1}^{m+n}\left[\binom{2j-2}{m-1}+\binom{2j-2}{n-1}\right]\zeta(2j-1)\zeta(m+n-2j+1)\label{2}\tag{2}$$
I know that \eqref{2} follows from \eqref{1} by using the well-known identity
$$\sum_{k=1}^\infty \frac{H_k^{(m)}}{k^n}+\sum_{k=1}^\infty \frac{H_k^{(n)}}{k^m}=\zeta(m)\zeta(n)+\zeta(m+n)$$
The proof of \eqref{1} may be found here but I am looking for different ones if possible. Thanks
Update (Nov 2 2023)
In this preprint, we proved the following generalizations for integers $p$ and $q$:
\begin{multline*}
i)\sum_{n=1}^\infty\frac{H_n^{(p)}}{n^q}=\frac12\zeta(p+q)-(-1)^q\sum_{k=0}^{\lfloor{\frac{q-1}{2}}\rfloor}\binom{p+q-2k-1}{p-1}\zeta(2k)\zeta(p+q-2k)\\
-(-1)^q\sum_{k=0}^{\lfloor{\frac{p}{2}}\rfloor}\binom{p+q-2k-1}{q-1}\zeta(2k)\zeta(p+q-2k);
\end{multline*}
\begin{multline*}
ii)\sum_{n=1}^\infty\frac{(-1)^nH_n^{(p)}}{n^q}=-\frac12\eta(p+q)+(-1)^q\sum_{k=0}^{\lfloor{\frac{q-1}{2}}\rfloor}\binom{p+q-2k-1}{p-1}\eta(2k)\zeta(p+q-2k)\\
-(-1)^q\sum_{k=0}^{\lfloor{\frac{p}{2}}\rfloor}\binom{p+q-2k-1}{q-1}\eta(2k)\eta(p+q-2k);
\end{multline*}
\begin{multline*}
iii)\sum_{n=1}^\infty\frac{\overline{H}_n^{(p)}}{n^q}=\frac12\eta(p+q)+\frac12(1+(-1)^{q})\zeta(q)\eta(p)\\
+(-1)^q\sum_{k=0}^{\lfloor{\frac{q}{2}}\rfloor}\binom{p+q-2k-1}{p-1}\eta(2k)\eta(p+q-2k)\\
-(-1)^q\sum_{k=0}^{\lfloor{\frac{p}{2}}\rfloor}\binom{p+q-2k-1}{q-1}\eta(2k)\zeta(p+q-2k);
\end{multline*}
\begin{multline*}iv)\sum_{n=1}^\infty\frac{(-1)^n \overline{H}_n^{(p)}}{n^q}=-\frac12\zeta(p+q)-\frac12(1+(-1)^q)\eta(q)\eta(p)\\
-(-1)^q\sum_{k=0}^{\lfloor{\frac{q}{2}}\rfloor}\binom{p+q-2k-1}{p-1}\zeta(2k)\eta(p+q-2k)\\
-(-1)^q\sum_{k=0}^{\lfloor{\frac{p}{2}}\rfloor}\binom{p+q-2k-1}{q-1}\zeta(2k)\eta(p+q-2k);
\end{multline*}
\begin{multline*}
v)\sum_{n=1}^\infty\frac{H_n^{(p)}}{(2n+1)^q}=-(1+(-1)^q)2^{p-1}\lambda(q)\eta(p)\\
-(-1)^q2^{p}\sum_{k=0}^{\lfloor{\frac{q-1}{2}}\rfloor}\binom{p+q-2k-1}{p-1}\lambda(2k)\lambda(p+q-2k)\\
-(-1)^q2^{p}\sum_{k=0}^{\lfloor{\frac{p}{2}}\rfloor}\binom{p+q-2k-1}{q-1}2^{-2k}\zeta(2k)\lambda(p+q-2k);
\end{multline*}
\begin{multline*}
vi)\sum_{n=1}^\infty\frac{\overline{H}_n^{(p)}}{(2n+1)^q}=\frac12(1+(-1)^q)\lambda(q)\eta(p)\\
+(-1)^q2^{p-1}\sum_{k=0}^{\lfloor{\frac{q-1}{2}}\rfloor}\binom{p+q-2k-2}{p-1}\frac{|E_{2k}|}{(2k)!}\left(\frac{\pi}{2}\right)^{2k+1}\beta(p+q-2k-1)\\
-(-1)^q2^{p}\sum_{k=0}^{\lfloor{\frac{p}{2}}\rfloor}\binom{p+q-2k-1}{q-1}2^{-2k}\eta(2k)\lambda(p+q-2k);
\end{multline*}
\begin{multline*}
vii)\sum_{n=1}^\infty \frac{(-1)^{n}H_n^{(p)}}{(2n+1)^q}=-(1-(-1)^q)2^{p-2}\frac{|E_{q-1}|}{(q-1)!}\left(\frac{\pi}{2}\right)^{q}\eta(p)\\
+(-1)^q2^{p-1}\sum_{k=0}^{\lfloor{\frac{q-2}{2}}\rfloor}\binom{p+q-2k-2}{p-1}\frac{|E_{2k}|}{(2k)!}\left(\frac{\pi}{2}\right)^{2k+1}\lambda(p+q-2k-1)\\
-(-1)^q2^{p}\sum_{k=0}^{\lfloor{\frac{p}{2}}\rfloor}\binom{p+q-2k-1}{q-1}2^{-2k}\eta(2k)\beta(p+q-2k);
\end{multline*}
\begin{multline*}
viii)\sum_{n=1}^\infty\frac{(-1)^n\overline{H}_n^{(p)}}{(2n+1)^q}=\frac{1-(-1)^q}{4(q-1)!}|E_{q-1}|\left(\frac{\pi}{2}\right)^q\eta(p)\\
-(-1)^q2^{p}\sum_{k=0}^{\lfloor{\frac{q}{2}}\rfloor}\binom{p+q-2k-1}{p-1}\lambda(2k)\beta(p+q-2k)\\
-(-1)^q2^{p}\sum_{k=0}^{\lfloor{\frac{p}{2}}\rfloor}\binom{p+q-2k-1}{q-1}2^{-2k}\zeta(2k)\beta(p+q-2k),
\end{multline*}
where $\zeta(s)$ is the Riemann zeta function, $\eta(s)=(1-2^{1-s})\zeta(s)$ is the Dirichlet eta function, $\lambda(s)=(1-2^{-s})\zeta(s)$ is the Dirichlet lambda function, $\beta$ is the Dirichlet beta function, and $E$ is the Euler number.
Note that $p\ge1$ and $q\ge2$ in the non-alternating sums, and $p,q\ge1$ in the alternating sums. We also have the sum weight $(p+q)$ is odd in the first six generalizations and even in the last two generalizations.
Best Answer
The following is the proof by Flajolet & Salvy* but with a lot more detail filled in which might make for a useful reference/post. Consider evaluating the following sum $${{S}_{p,q}}=\sum\limits_{n=1}^{\infty }{\frac{H_{n}^{p}}{{{n}^{q}}}}$$ where $$H_{n}^{p}=\sum\limits_{k=1}^{n}{\frac{1}{{{k}^{p}}}}$$ are the generalised harmonic numbers and where $p+q=odd$. To do this we will need the following local expansions $$\begin{aligned}\frac{1}{{{z}^{q}}}&\underset{z\to n}{\mathop{=}}\,\sum\limits_{k=0}^{\infty }{{{\left( -1 \right)}^{k}}\left( \begin{matrix}q-1+k \\ q-1 \\ \end{matrix} \right)\frac{{{\left( z-n \right)}^{k}}}{{{n}^{q+k}}}\ \ \ n\ne 0,q\in \mathbb{Z}} \\\pi\cot \left( \pi z \right)&\underset{z\to n}{\mathop{=}}\,\frac{1}{z-n}-2\sum\limits_{k=1}^{\infty }{\zeta \left( 2k \right){{\left( z-n \right)}^{2k-1}}}\\\frac{{{\psi }^{p-1}}\left( -z \right)}{\left( p-1 \right)!}&\underset{z\to n}{\mathop{=}}\,\frac{1}{{{\left( z-n \right)}^{p}}}\left( 1+{{\left( -1 \right)}^{p}}\sum\limits_{k\ge p}^{\infty }{\left( \begin{matrix} k-1 \\ p-1 \\ \end{matrix} \right)\left( \zeta \left( k \right)+{{\left( -1 \right)}^{k}}H_{n}^{k} \right){{\left( z-n \right)}^{k}}} \right)\ \ \ n\ge 0,p>1\\\frac{{{\psi }^{p-1}}\left( -z \right)}{\left( p-1 \right)!}&\underset{z\to -n}{\mathop{=}}\,{{\left( -1 \right)}^{p}}\sum\limits_{k\ge 0}^{\infty }{\left( \begin{matrix} p-1+k \\ p-1 \\ \end{matrix} \right)\left( \zeta \left( p+k \right)-H_{n-1}^{p+k} \right){{\left( z+n \right)}^{k}}}\ \ \ n>0,p>1 \end{aligned}$$
Now consider, for $q>1$, $$\frac{1}{2}\pi \cot \left( \pi z \right)\frac{{{\psi }^{\left( p-1 \right)}}\left( -z \right)}{\left( p-1 \right)!{{z}^{q}}}$$ which has poles at the integers and vanishes over a large enough contour. The sum over all residues of this function is zero. So consider therefore $z$ at the negative integers where we have simple poles: $$\begin{align}&\frac{1}{2}\pi \cot \left( \pi z \right)\frac{{{\psi }^{\left( p-1 \right)}}\left( -z \right)}{\left( p-1 \right)!{{z}^{q}}}\underset{z=-n}{\mathop{=}}\,\\&\frac{1}{2}{{\left( -1 \right)}^{p}}\frac{1}{{{n}^{q-1}}}\sum\limits_{k=0}^{\infty }{\left( \begin{matrix} p-1+k \\ p-1 \\ \end{matrix} \right)\left( \zeta \left( p+k \right)-H_{n-1}^{p+k} \right){{\left( z+n \right)}^{k}}}\left\{ \frac{1}{z+n}-2\sum\limits_{k=1}^{\infty }{\zeta \left( 2k \right){{\left( z+n \right)}^{2k-1}}} \right\}\end{align}$$ So at negative integers we have the residues $$\underset{z=-n}{\mathop{res}}\,\frac{1}{2}\pi \cot \left( \pi z \right)\frac{{{\psi }^{\left( p-1 \right)}}\left( -z \right)}{\left( p-1 \right)!{{z}^{q}}}=\frac{{{\left( -1 \right)}^{p+q}}}{2{{n}^{q}}}\left( \zeta \left( p \right)-H_{n-1}^{p} \right)$$ At positive integers we have poles of order $p+1$, and so $$\begin{align} \frac{1}{2}\pi \cot \left( \pi z \right)\frac{{{\psi }^{\left( p-1 \right)}}\left( -z \right)}{\left( p-1 \right)!{{z}^{q}}}&\underset{z=n}{\mathop{=}}\,\frac{1}{2}\sum\limits_{k=0}^{\infty }{{{\left( -1 \right)}^{k}}\left( \begin{matrix} q-1+k \\ q-1 \\ \end{matrix} \right)\frac{{{\left( z-n \right)}^{k-p-1}}}{{{n}^{q+k}}}}\\&-\sum\limits_{j=0}^{\infty }{{{\left( -1 \right)}^{j}}\left( \begin{matrix} q-1+j \\ q-1 \\ \end{matrix} \right)\frac{1}{{{n}^{q+j}}}}\sum\limits_{k=1}^{\infty }{\zeta \left( 2k \right){{\left( z-n \right)}^{2k-p+j-1}}} \\ & +\frac{\left( {{\left( -1 \right)}^{p}}\zeta \left( p \right)+H_{n}^{p} \right)}{2{{n}^{q}}\left( z-n \right)}+O\left( 1 \right) \\ \end{align}$$
We have therefore $$\begin{align}\underset{z=n}{\mathop{res}}\,\frac{1}{2}\pi \cot \left( \pi z \right)\frac{{{\psi }^{\left( p-1 \right)}}\left( -z \right)}{\left( p-1 \right)!{{z}^{q}}}&=\frac{1}{2{{n}^{q}}}\left( {{\left( -1 \right)}^{p}}\zeta \left( p \right)+H_{n}^{p} \right)+\left( \begin{matrix} p+q-1 \\ q-1 \\ \end{matrix} \right)\frac{{{\left( -1 \right)}^{p}}}{2{{n}^{p+q}}}\\&-{{\left( -1 \right)}^{p}}\sum\limits_{k=1}^{\left\lfloor p/2 \right\rfloor }{\left( \begin{matrix} p+q-2k-1 \\ q-1 \\ \end{matrix} \right)\frac{\zeta \left( 2k \right)}{{{n}^{p+q-2k}}}}\end{align}$$ There is a pole of order $q+p+1$ at $z=0$, i.e. $$\frac{1}{2}\pi \cot \left( \pi z \right)\frac{{{\psi }^{\left( p-1 \right)}}\left( -z \right)}{\left( p-1 \right)!{{z}^{q}}}\underset{z=0}{\mathop{=}}\,\frac{1}{2{{z}^{p+q}}}\left\{ \frac{1}{z}-2\sum\limits_{k=1}^{\infty }{\zeta \left( 2k \right){{z}^{2k-1}}} \right\}\left\{ 1+{{\left( -1 \right)}^{p}}\sum\limits_{k\ge p}^{{}}{\left( \begin{matrix} k-1 \\ p-1 \\ \end{matrix} \right)\zeta \left( k \right){{z}^{k}}} \right\}$$ So we find $$\begin{align} \frac{1}{2}\pi \cot \left( \pi z \right)\frac{{{\psi }^{\left( p-1 \right)}}\left( -z \right)}{\left( p-1 \right)!{{z}^{q}}}&\underset{z=0}{\mathop{=}}\,\frac{1}{2{{z}^{p+q+1}}}-\frac{1}{{{z}^{p+q}}}\sum\limits_{k=1}^{\infty }{\zeta \left( 2k \right){{z}^{2k-1}}} +\frac{{{\left( -1 \right)}^{p}}}{2{{z}^{p+q+1}}}\sum\limits_{k\ge p}^{\infty }{\left( \begin{matrix} k-1 \\ p-1 \\ \end{matrix} \right)\left( \zeta \left( k \right) \right){{z}^{k}}}\\&-\frac{{{\left( -1 \right)}^{p}}}{{{z}^{p+q}}}\sum\limits_{k=1}^{\infty }{\zeta \left( 2k \right){{z}^{2k-1}}}\sum\limits_{j=0}^{\infty }{\left( \begin{matrix} p+j-1 \\ p-1 \\ \end{matrix} \right)\left( \zeta \left( p+j \right) \right){{z}^{p+j}}} \\ \end{align}$$ Now because $p+q$ is odd the second term will contribute nothing to the residue. We find then
$$\begin{align}\underset{z=0}{\mathop{res}}\,\frac{1}{2}\pi \cot \left( \pi z \right)\frac{{{\psi }^{\left( p-1 \right)}}\left( -z \right)}{\left( p-1 \right)!{{z}^{q}}}&=\frac{{{\left( -1 \right)}^{p}}}{2}\left( \begin{matrix} p+q-1 \\ p-1 \\ \end{matrix} \right)\zeta \left( p+q \right)\\&-{{\left( -1 \right)}^{p}}\sum\limits_{k=1}^{\left\lfloor q/2 \right\rfloor }{\left( \begin{matrix} q+p-2k-1 \\ p-1 \\ \end{matrix} \right)\zeta \left( 2k \right)\zeta \left( q+p-2k \right)}\end{align}$$
Now the sum of all residues is zero. So we have then $\sum\limits_{n=1}^{\infty }{\underset{z\ne 0}{\mathop{res}}\,}=-\underset{z=0}{\mathop{res}}\,$, hence $$\begin{align} & \sum\limits_{n=1}^{\infty }{{}}\frac{{{\left( -1 \right)}^{p+q}}}{2{{n}^{q}}}\left( \zeta \left( p \right)-H_{n-1}^{p} \right) +\frac{1}{2{{n}^{q}}}\left( {{\left( -1 \right)}^{p}}\zeta \left( p \right) + H_{n}^{p} \right)\\&+\left( \begin{matrix} p+q-1 \\ q-1 \\ \end{matrix} \right)\frac{{{\left( -1 \right)}^{p}}}{2{{n}^{p+q}}}-{{\left( -1 \right)}^{p}}\sum\limits_{k=1}^{\left\lfloor p/2 \right\rfloor }{\left( \begin{matrix} p+q-2k-1 \\ q-1 \\ \end{matrix} \right)\frac{\zeta \left( 2k \right)}{{{n}^{p+q-2k}}}} \\ & =-\frac{{{\left( -1 \right)}^{p}}}{2}\left( \begin{matrix} p+q-1 \\ p-1 \\ \end{matrix} \right)\zeta \left( p+q \right)+{{\left( -1 \right)}^{p}}\sum\limits_{k=1}^{\left\lfloor q/2 \right\rfloor }{\left( \begin{matrix} q+p-2k-1 \\ p-1 \\ \end{matrix} \right)\zeta \left( 2k \right)\zeta \left( q+p-2k \right)} \\ \end{align}$$ Summing over $n$ $$\begin{align} \sum\limits_{n=1}^{\infty }{\frac{H_{n}^{p}}{{{n}^{q}}}} & =\left\{ 1-{{\left( -1 \right)}^{p}}\left( \begin{matrix} p+q-1 \\ p-1 \\ \end{matrix} \right)-{{\left( -1 \right)}^{p}}\left( \begin{matrix} p+q-1 \\ q-1 \\ \end{matrix} \right) \right\}\frac{1}{2}\zeta \left( p+q \right)+\frac{1-{{\left( -1 \right)}^{p}}}{2}\zeta \left( p \right)\zeta \left( q \right) \\ & +{{\left( -1 \right)}^{p}}\sum\limits_{k=1}^{\left\lfloor q/2 \right\rfloor }{\left( \begin{matrix} q+p-2k-1 \\ p-1 \\ \end{matrix} \right)\zeta \left( 2k \right)\zeta \left( q+p-2k \right)} \\ & +{{\left( -1 \right)}^{p}}\sum\limits_{k=1}^{\left\lfloor p/2 \right\rfloor }{\left( \begin{matrix} p+q-2k-1 \\ q-1 \\ \end{matrix} \right)\zeta \left( 2k \right)\zeta \left( p+q-2k \right)} \\ \end{align}$$ * “Euler Sums and Contour Integral Representations” Phillipe Flajolet & Bruno Salvy, Experimental Mathematics, Vol. 7 (1998), No. 1