Evaluating sums of the form $\sum_{n=1}^{\infty}\frac{1}{n^a}\sum_{m=1}^{n}\frac{1}{m^b}$

Question

How to tackle infinite sums of the type $$S(a,b)=\sum_{n=1}^{\infty}\frac{1}{n^a}\sum_{m=1}^{n}\frac{1}{m^b}$$ where $a$ and $b$ are natural numbers. Zeta function is involved, for example $S(3,2)=3\zeta(2)\zeta(3)-\frac{9}{2}\zeta(5)$, but so far I can't figure out how to relate them.

Answer 1

According to my knowledge there is no general formula for

$$S(p,q) = \sum_{n=1}^\infty\frac{H^{(p)}}{n^q}$$

Where

$$H_n^{(p)} = \sum_{k=1}^n \frac{1}{k^p}$$

We have some formulas for special cases

• If $p$ is even and $q$ is odd or the other way around see here.
• If $p=1$ we have

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

The proof Generalized Euler sum $\sum_{n=1}^\infty \frac{H_n}{n^q}$

• Symmetric formula

$$S(p,q)+S(q,p) = \zeta(p+q)+\zeta(p)\zeta(q)$$

See the proof here page 106

• For $p=q=k$

$$S(k,k) = \frac{\zeta(2k)+\zeta^2(k)}{2}$$

The proof Intruiging Symmetric harmonic sum $\sum_{n\geq 1} \frac{H^{(k)}_n}{n^k}\, = \frac{\zeta{(2k)}+\zeta^{2}(k)}{2}$

• For small order $p,q$ Usually we consider the generating function

$$\sum_{n=1}^\infty H^{p}_k x^k = \frac{\mathrm{Li}_p(x)}{1-x}$$

See the proof here , page [97] .

• Relate it to integration through

$$S(p,q) = \zeta(p)\zeta(q) - \frac{(-1)^{p-1}}{(p-1)!}\int^1_0 \frac{\log^{p-1}(x) \mathrm{Li}_q(x)}{1-x}\,dx$$

The proof is page 4 in PEDRO FREITAS paper on integrals of polygarithmic functions here.

• There also exists a complex analysis approach usually for

$$S_{p^r,q}\sum_{n=1}^\infty \frac{(H_n^{(p)})^r}{n^q}$$

by Philippe Flajolet and Bruno Salvy here.