$\zeta(4)$ in terms of a series of $\zeta(3)$ and harmonic numbers The other day I believe I found a proof that:
$$\sum_{k=1}^\infty \frac{\zeta(2)-H_k^{(2)}}{k} = \zeta(3)$$
I was wondering if a general recursion like this was well known, but I couldn't find anything.
I tried the following result which seems to hold.
$$\sum_{k=1}^\infty \frac{\zeta(3)-H_k^{(3)}}{k} = \frac{\zeta(4)}{4}$$
Can anyone prove that this last result is correct or incorrect? Thanks.
 A: We have $$\zeta(a)-H_n^{(a)}=\frac{(-1)^{a-1}}{(a-1)!}\int_0^1 \frac{x^n\ln^{a-1}(x)}{1-x}dx$$
Then
$$\sum_{n=1}^\infty\frac{\zeta(a)-H_n^{(a)}}{n}=\frac{(-1)^{a-1}}{(a-1)!}\int_0^1 \frac{\ln^{a-1}(x)}{1-x}\left(\sum_{n=1}^\infty\frac{x^n}{n}\right)dx$$
$$\frac{(-1)^{a-1}}{(a-1)!}\int_0^1 \frac{-\ln^{a-1}(x)\ln(1-x)}{1-x}dx=\frac{(-1)^{a-1}}{(a-1)!}\sum_{n=1}^\infty H_n\int_0^1x^{n}\ln^{a-1}(x)dx$$
$$=\frac{(-1)^{a-1}}{(a-1)!}\sum_{n=1}^\infty H_n\left(\frac{(-1)^{a-1}(a-1)!}{(n+1)^a}\right)=\sum_{n=1}^\infty\frac{H_n}{(n+1)^a}$$
$$=\sum_{n=1}^\infty\frac{H_n}{n^a}-\zeta(a+1)$$
We have the generalized Euler sum
$$\sum_{n=1}^\infty\frac{H_n}{n^a}
=\frac{a+2}{2}\zeta(a+1)-\frac12\sum_{j=1}^{a-2}\zeta(a-j)\zeta(j+1)$$
Thus
$$\boxed{\sum_{n=1}^\infty\frac{\zeta(a)-H_n^{(a)}}{n}=\frac{a}{2}\zeta(a+1)-\frac12\sum_{j=1}^{a-2}\zeta(a-j)\zeta(j+1)}.$$

An alternative way using Abel's summation:
$$\sum_{k=1}^n a_k b_k=b_{n}A_n+\sum_{k=1}^{n-1}A_k\left(b_{k}-b_{k+1}\right)$$
where $ A_n=\sum_{i=1}^n a_i$
Let $b_k=\zeta(a)-H_k^{(a)}$ and $a_k=\frac1k$ we have
$$\sum_{k=1}^n \frac{\zeta(a)-H_k^{(a)}}{k}=(\zeta(a)-H_n^{(a)})\sum_{i=1}^n\frac1i+\sum_{k=1}^{n-1}\left(\sum_{i=1}^k\frac1i\right)\left(-H_{k}^{(a)}+H_{k+1}^{(a)}\right)$$
$$=(\zeta(a)-H_n^{(a)})H_n+\sum_{k=1}^{n-1}\left(H_k\right)\left(\frac{1}{(k+1)^a}\right)$$
Let $n\mapsto \infty$ we have
$$\sum_{k=1}^\infty \frac{\zeta(a)-H_k^{(a)}}{k}=0+\sum_{k=1}^\infty\frac{H_k}{(k+1)^a}=\sum_{k=1}^\infty\frac{H_k}{k^a}-\zeta(a+1)$$
