TITLE: Generalized Euler sum $\sum_{n=1}^\infty \frac{H_n}{n^q}$
QUESTION [80 upvotes]: 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 
$$\sum_{n=1}^\infty (-1)^{n+1}\frac{H_n}{n^q}$$

REPLY [2 votes]: Partial solution:
I am going to prove 

$$\sum_{k=1}^\infty\frac{H_k}{k^n}=\frac12\sum_{i=1}^{n-2}(-1)^{i-1}\zeta(n-i)\zeta(i+1),\quad n=3,5,7, ...$$


We have 
$$\int_0^1x^{k-1}\operatorname{Li}_n(x)\ dx\overset{IBP}{=}(-1)^{n-1}\frac{H_k}{k^n}-\sum_{i=1}^{n-1}(-1)^i\frac{\zeta(n-i+1)}{k^i}$$
Divide both sides by $k$ then consider the summation from $k=1$ to $\infty$ we have
$$\int_0^1\frac{\operatorname{Li}_n(x)}{x}\sum_{k=1}^\infty\frac{x^k}{k}\ dx=(-1)^{n-1}\sum_{k=1}^\infty\frac{H_k}{k^{n+1}}-\sum_{i=1}^{n-1}(-1)^i\zeta(n-i+1)\sum_{k=1}^\infty\frac1{k^{i+1}}$$
$$\small{-\int_0^1\frac{\operatorname{Li}_n(x)\ln(1-x)}{x}\ dx=(-1)^{n-1}\sum_{k=1}^\infty\frac{H_k}{k^{n+1}}-\sum_{i=1}^{n-1}(-1)^i\zeta(n-i+1)\zeta(i+1)}\tag1$$
where  
$$-\int_0^1\frac{\operatorname{Li}_n(x)\ln(1-x)}{x}\ dx=-\sum_{k=1}^\infty\frac1{k^n}\int_0^1 x^{k-1}\ln(1-x)\ dx=\sum_{k=1}^\infty\frac{H_k}{k^{n+1}}\tag2$$
Plug (2) in (1) we get
$$\sum_{k=1}^\infty\frac{H_k}{k^{n+1}}[1+(-1)^n]=-\sum_{i=1}^{n-1}(-1)^i\zeta(n-i+1)\zeta(i+1)$$
Let $n-1\mapsto n$ to get
$$\sum_{k=1}^\infty\frac{H_k}{k^{n}}[1-(-1)^n]=-\sum_{i=1}^{n-2}(-1)^i\zeta(n-i)\zeta(i+1)$$
So clearly for odd $n\geq3$ we have 
$$\sum_{k=1}^\infty\frac{H_k}{k^{n}}[2]=-\sum_{i=1}^{n-2}(-1)^i\zeta(n-i)\zeta(i+1)$$
set $n=2m+1$
$$\sum_{k=1}^\infty\frac{H_k}{k^{2m+1}}=-\frac12\sum_{i=1}^{2m-1}(-1)^i\zeta(2m+1-i)\zeta(i+1),\quad m=1,2,3,...$$