# Closed form of Euler-type sum over zeta functions $\sum _{k=2}^{\infty } \frac{\zeta (k)}{k^2}$?

Revisiting the question on the integral over the harmonic number I stumbled over the nice formula

$$\sum_{k\ge2} (-1)^{k+1}\frac{\zeta(k)}{k} = \gamma\tag{1}$$

where $$\zeta(z)$$ is the Riemann zeta function and $$\gamma$$ is Euler's gamma.

Searching SE I found solutions to related but even more complicated problems (see below) so I dropped $$(1)$$ and propose here instead the problem asked in the heading, viz. to find a closed expression for

$$s=\sum_{k\ge2} \frac{\zeta(k)}{k^2} \simeq 0.835998 \tag{2}$$

I tried several approaches but still couldn't find a closed expression. So I would consider this a tough sum.

More generally we can ask for sums of the form

$$s_q=\sum_{k\ge2} \frac{\zeta(k)}{k^q} \tag{3}$$

A similar tough sums is

$$s_{-1}=\sum_{k\ge2} \frac{\zeta(k)}{k(k-1)} = ? \tag{4}$$

Whereas for sums of the type $$\sum_{k\ge2} \frac{\zeta(k)}{k(k+1)}$$, $$\sum_{k\ge2} \frac{\zeta(k)}{k(k+1)(k+2)}$$, or generally

$$s_{p(m)}=\sum_{k\ge2} \frac{\zeta(k)}{(k)_m}=\text{closed expression} \tag{5}$$

where $$(k)_m=k(k+1)(k+2)\ldots (k+m-1)$$ is the Pochhammer symbol, closed expressions for any integer $$m\ge2$$ can be found using CAS (e.g. Mathematica).

My effort so far

Because of the length of these developments I have put them into a (preliminary) self answer.

Related problems

• Since $$\sum_{k\geq 2}\zeta(k)\frac{x^k}{k} = -\gamma x+\log\Gamma(1-x)$$ we have $$\sum_{k\geq 2}\frac{\zeta(k)}{k^2}=-\gamma+\int_{0}^{1}\frac{\log\Gamma(1-x)}{x}\,dx\stackrel{\text{IBP}}{=}-\gamma+\int_{0}^{1}\psi(x)\log(1-x)\,dx$$ so it might be useful to have closed forms for $$\int_{0}^{1}x^m \psi(x)\,dx.$$ – Jack D'Aurizio Mar 5 '20 at 16:19
• @Jack D'Aurizio I've thought about that, but the integral is even more complicated (see my answer (A.3)). – Dr. Wolfgang Hintze Mar 5 '20 at 17:04

This is a preliminary incomplete answer showing my effort to solve the problem.

What I did so far to find a closed expression to $$(2)$$ is mainly a reformulation. Maybe someone recognises one of these expressions.

It turned out that different approaches sometimes lead to the same result. I have therefore indicated "really" different formulas by putting them in a box.

0) Just be reassured not to miss trivial things I consulted the online-encyclopedia of integer sequences with the first few digits of $$N(s)$$.

Nothing relevant was found but the first 5 digits appear somewhere in several funny numbers, like https://oeis.org/A019694, Decimal expansion of 2*Pi/5.

1) Expanding zeta in a series and changing the order of summation, leaves another sum

$$s=\sum _{k=2}^{\infty } \frac{\zeta (k)}{k^2}=\sum _{k=1}^{\infty } \frac{1}{k^2}\sum _{m=1}^{\infty } \frac{1}{m^k} \\ =\sum _{m=1}^{\infty } \left(\sum _{k=2}^{\infty } \frac{1}{k^2 m^k}\right) \\ \boxed{s=\sum _{m=1}^{\infty } \left(\operatorname{Li}_2\left(\frac{1}{m}\right)-\frac{1}{m}\right)}\tag{A.1}$$

2) Replacing inverse power $$\frac{1}{k^2}$$ by an integral and doing the sum, leaves a nice compact integral

In fact,

$$\int_0^1 x^{k-1} \log \left(\frac{1}{x}\right) \, dx=\frac{1}{k^2}\tag{A.2.1}$$

and using the Taylor expansion of the harmonic number

$$\sum _{k=2}^{\infty } x^{k-1} \zeta (k)=-H_{-x}\tag{A.2.2}$$

we have

$$s=\sum _{k=2}^{\infty } \zeta (k) \int_0^1 x^{k-1} \log \left(\frac{1}{x}\right) \, dx \\ =\int_0^1 \log \left(\frac{1}{x}\right) \sum _{k=2}^{\infty } x^{k-1} \zeta (k) \, dx=\int_0^1 \left(-H_{-x}\right) \log \left(\frac{1}{x}\right) \, dx \\ \boxed{s=\int_0^1 H_{-x} \log (x) \, dx}\tag{A.2.3}$$

3) Exploring the integral $$\int_0^1 H_{-x} \log (x) \, dx$$

EDIT 06.03.20 begin

Using the basic relation $$H_n=H_{n-1}+\frac{1}{n}$$ and letting $$n=1-x$$ we can replace $$H_{-x} \to H_{1-x} -\frac{1}{1-x}$$ which, observing $$\int_0^1\frac{\log(x)}{1-x}\,dx = -\zeta(2)$$, leads to the possibly more pleasant form

$$s = \zeta(2) + \int_0^1 \log(1-x) H_{x}\,dx\tag{A.3.0}$$

EDIT end

Integrating by parts, $$\int H_{-x} \, dx=\gamma x-\text{log\Gamma }(1-x)$$, gives

$$s=\int_0^1 H_{-x} \log (x) \, dx=\int_0^1 \frac{(\operatorname{\log\Gamma}(1-x)-x\gamma) }{x} \, dx\tag{A.3}$$

Here it might appear helpful to have the generating integral

$$s(\xi)=\int_0^1 x^\xi H_{-x} \, dx\tag{A.3.1}$$

so that we can generate the $$\log$$ by the derivative with respect to $$\xi$$. But that integral is divergent at $$x=1$$.

3a) Inserting the definition of $$H$$ as an integral leaves another integral

$$s=\int_0^1 \left(\int_0^1 \frac{\left(1-z^{-x}\right) \log (x)}{1-z} \, dz\right) \, dx \\ =\int_0^1 \left(\int_0^1 \frac{\left(1-z^{-x}\right) \log (x)}{1-z} \, dx\right) \, dz \\ =\int_0^1 \frac{-\log (z)+\log (\log (z))+\Gamma (0,\log (z))+\gamma }{\log (z)-z \log (z)} \, dz \\ \boxed{s=\int_0^{\infty } \frac{t+\log (-t)+\Gamma (0,-t)+\gamma }{t \left(1-e^{t}\right)} \, dt}\tag{A.3.2}$$

3b) Inserting the definition of $$H$$ as an infinite sum, leaves another infinite sum

$$s=\int_0^1 \log (x) \sum _{m=1}^{\infty } \left(\frac{1}{m}-\frac{1}{m-x}\right) \, dx \\ =\sum _{m=1}^{\infty } \int_0^1 \left(\frac{1}{m}-\frac{1}{m-x}\right) \log (x) \, dx=\sum _{m=1}^{\infty } c(m)\tag{A.3.3}$$

with

$$c(1)=\frac{1}{6} \left(\pi ^2-6\right)\tag{A.3.4}$$

and

$$c(m\gt1)=\int_0^1 \left(\frac{1}{m}-\frac{1}{m-x}\right) \log (x) \, dx \\ =-\operatorname{Li}_2\left(\frac{m-1}{m}\right)-\frac{1}{m}-\log ^2(m)+\log (m-1) \log (m)+\frac{\pi ^2}{6}\tag{A.3.5}$$

This can be simplified appreciable using the transformation formula

$$\text{Li}_2\left(\frac{m-1}{m}\right)=-\text{Li}_2\left(\frac{1}{m}\right)-\log \left(\frac{1}{m}\right) \log \left(\frac{m-1}{m}\right)+\frac{\pi ^2}{6}$$

to give

$$c(m) = \text{Li}_2\left(\frac{1}{m}\right)-\frac{1}{m}$$

so that we have found a complicated way to regain exactly $$(A.1)$$.

4) Replace zeta by an integral, leaves another integral

We have

$$\zeta (k)=\frac{1}{\Gamma (k)}\int_0^{\infty } \frac{t^{k-1}}{e^t-1} \, dt\tag{A.4.1},$$

so that our sum becomes

$$s=\sum_{k\ge2} \frac{1}{k^2}\frac{1}{\Gamma (k)}\int_0^{\infty } \frac{t^{k-1}}{e^t-1} \, dt=\int_0^{\infty } \frac{1}{e^t-1}\left( \sum_{k\ge2}\frac{1}{k^2}\frac{t^{k-1}}{\Gamma (k)}\right)\, dt \\ =\int_0^{\infty } \frac{-\log (-t)-\Gamma (0,-t)-e^t \Gamma (2,t)-\gamma +1}{t \left(e^t-1\right)} \, dt\tag{A.4.2}$$

We can simplify the integrand.

The incomplete gamma function is defined as

$$\Gamma (r,y)=\int_y^{\infty }x^{r-1} \exp (-x)\, dx\tag{A.4.3}$$

This gives

$$\Gamma (2,t) =e^{-t} (t+1)\tag{A.4.3a}$$

and we can see (e.g. by plotting) that the combination

$$-\log (-t)-\Gamma (0,-t)\tag{A.4.3b}$$

is real for all real $$t$$. I don't know a name for this expression.

This gives finally

$$s = \int_0^{\infty } \frac{1}{t(1-e^t)} \left(t+\log (-t)+\Gamma (0,-t)+\gamma \right)\, dt\tag{A.4.4}$$

which coincides with the last formula of $$(A.3.2)$$.

5) Generating functions

Defining the generating functions analogous to $$(3)$$

$$g(q,z) =\sum_{k\ge2} \frac{z^k}{k^q}\zeta(k)\tag{A.5.1}$$

we have

$$g(0,z) = -z (\psi ^{(0)}(1-z)+\gamma ) = - z H_{-z}\tag{A.5.2}$$

and the sequence

$$g(q,z)=\int_{0}^z \frac{g(q-1,y)}{y}\,dy, q=1,2,\ldots \tag{A.5.3}$$

Giving

$$g(1,z)=\int_0^z H_{-y} \, dy=\gamma z-\operatorname{\log\Gamma}(1-z)\tag{A.5.4}$$

and the g.f. we are looking for

$$g(2,z)=\gamma z-\int_0^z \frac{1}{y}\operatorname{\log\Gamma}(1-y) \, dy=\text{?}\tag{A.5.5}$$

This one we have encountered already in $$(A.3)$$.

Notice that interestingly $$\lim_{z\to -1} \, g(0,z)=1$$ in spite of the fact, that the series diverges. In fact, there is no limit but two partial sums with even and odd parity tend to $$\frac{1}{2}$$ and $$\frac{3}{2}$$, respectively, i.e. the sequence has two accumulation points, and their arithmetic mean is $$=1$$.

6) Complex contour integral

I am not sure if this approach could lead to a closed expression but it might be interesting.

Representing the infinite series as a complex contour integral with the "kernel function" $$H_{-z}$$ and a path coming from $$i+\infty$$, going to $$i+\frac{3}{2}$$, to $$-i+\frac{3}{2}$$, and then back to $$-i+\infty$$, then bending the path around we arrive at the following representation of our sum $$s$$:

$$s=2-\gamma -\frac{1}{2 \pi }\int_{\frac{1}{2}-i\infty }^{\frac{1}{2}+i \infty } \frac{H_{-z} \zeta \left(z\right)}{z^2} \, dz\tag{6.1}$$

where the terms before the integral are the residue of the integrand at $$z=1$$:

$$\text{Res}\left(\frac{H_{-z} \zeta (z)}{z^2}\right)|_{z=1} =-2 + \gamma\tag{6.2}$$

Notice that the integral is taken on the critical strip where the zeta function has its non-trivial zeroes (if Riemann was right).