The following series expression:
$$f(s) = \sum_{n=2}^\infty \frac{\zeta(n)-1}{n^s} \tag{1}$$
has well known closed forms for $s=0$ and $s=1$, however also seems to converge for $\Re(s) < 0$. After some experimentation I conjecture that:
$$f(-m) = 1+ \sum_{k=1}^m k!\, \mathrm{Stirling2}(m+1,k+1)\, \zeta(k+1) \tag{2}$$
for $m = 0, 1, 2, 3, …$ with Stirling2 a Stirling number of the second kind.
Similarly for the alternating series:
$$g(s) = \sum_{n=2}^\infty (-1)^n\,\frac{\zeta(n)-1}{n^s} \tag{3}$$
I conjecture:
$$g(-m) = -\frac{E_m(2)}{2} – \sum_{k=1}^m (-1)^k \,k!\, \mathrm{Stirling2}(m+1,k+1)\, \zeta(k+1) \tag{4}$$
for $m = 1, 2, 3,…$. With $E_m(2)$ the m-th Euler Polynomial evaluated at $2$.
Is this indeed true?
Best Answer
We can use the formula for the polylogarithm $\operatorname{Li}_{-m}(z)$ where $m$ is a nonnegative integer.
$$ \begin{align} \sum_{n=2}^{\infty} n^{m} \left( \zeta(n)-1 \right) &= \sum_{n=2}^{\infty} n^{m} \sum_{k=2}^{\infty} \frac{1}{k^{n}} \\ &= \sum_{k=2}^{\infty} \sum_{n=2}^{\infty} \frac{n^{m}}{k^{n}} \\ &= \sum_{k=2}^{\infty} \left( \operatorname{Li}_{-m} \left(\frac{1}{k} \right) -\frac{1}{k} \right) \\ &= \sum_{k=2}^{\infty} \left(\sum_{j=0}^{m} j! S (m+1,j+1 ) \left(\frac{1}{k-1}\right)^{j+1}-\frac{1}{k}\right) \\ &= \sum_{k=2}^{\infty} \left(\sum_{j=\color{red}{1}}^{m} j! S (m+1,j+1 ) \left(\frac{1}{k-1}\right)^{j+1} + \frac{1}{k-1}-\frac{1}{k}\right) \\ &= \sum_{k=2}^{\infty} \sum_{j=1}^{m}j! S (m+1,j+1 ) \left(\frac{1}{k-1}\right)^{j+1} + \sum_{k=2}^{\infty} \left(\frac{1}{k-1}-\frac{1}{k} \right) \\ &= \sum_{j=1}^{m} \sum_{k=2}^{\infty}j! S (m+1,j+1 ) \left(\frac{1}{k-1}\right)^{j+1} + 1 \\ &= \sum_{j=1}^{m}j! S (m+1,j+1 ) \zeta(j+1)+1 \end{align}$$
Using the same approach for the alternating version, I get $$\sum_{n=2}^{\infty} (-1)^{n} n^{m} \left(\zeta(n)-1 \right) = \sum_{j=1}^{m} (-1)^{j+1} j! S(m+1,j+1) \left(\zeta(j+1)-1-\frac{1}{2^{j+1}} \right) + \frac{1}{2},$$ which seems to give correct values.