$\zeta(4)$ 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.

1 Answer

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

$$\sum_{k=1}^\infty \frac{\zeta(a)-H_k^{(a)}}{k}=0+\sum_{k=1}^\infty\frac{H_k}{(k+1)^a}=\sum_{k=1}^\infty\frac{H_k}{k^a}-\zeta(a+1)$$

• Such strategies appear in the book, (Almost) Impossible Integrals, Sums, and Series. For example, at pages $479$-$481$ it is calculated the version $\displaystyle \sum_{n=1}^{\infty}\frac{H_n}{n}\left(\zeta(3)-1-1/2^3-...-1/n^3\right)$ which can also be written as $\displaystyle \sum_{n=1}^{\infty}\frac{H_n}{n}\left(\zeta(3)-H_n^{(3)}\right)$. – user97357329 Jul 12 at 8:13
• Also (+1) for the generalizations. – user97357329 Jul 12 at 8:32
• @user97357329 thank you. Yes that book is full of powerful identities. – Ali Shather Jul 12 at 8:34