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As per this question, I'm trying to evaluate $$\sum_{k=2}^{\infty} \big{(} \zeta(n)^{2}-1 \big{)} = -\sum_{m=2}^{\infty} \frac{H_{-\frac{1}{m}}}{m}. $$

Here, $H_{x}$ is a generalized Harmonic number. When we proceed with the definition $$\begin{align} H_{x} = \sum_{k=1}^{\infty} \binom{x}{k} \frac{(-1)^{k}}{k} \end{align} $$ and consider the fact that relates this expression to the unsigned Stirling numbers of the first kind: $$\begin{align} (-1)^n {-m \choose n} = {n+m-1 \choose n} = \frac{1}{n!} \sum_{i=0}^n \left[ {n \atop i} \right] m^i , \end{align} $$ we can replace $m$ by $\frac{1}{m}$ and rearrange sums to obtain:

$$\begin{align} \sum_{m=2}^{\infty} \frac{H_{-\frac{1}{m}}}{m} &= - \sum_{m=2}^{\infty} \sum_{k=1}^{\infty} \sum_{i=0}^{k} \frac{1}{k k!} m^{-(i+1)} \left[ {k \atop i} \right] \\ &= - \sum_{k=1}^{\infty} \sum_{i=0}^{k} \frac{1}{k k!} \left[ {k \atop i} \right] \sum_{m=2}^{\infty} m^{-(i+1)} \\ &= - \sum_{k=1}^{\infty} \sum_{i=0}^{k} \frac{1}{k k!} \left[ {k \atop i} \right] (\zeta(i+1) - 1) \\ &= - \bigg{(} \sum_{k=1}^{\infty} \frac{1}{kk!} \bigg{(} \sum_{i=0}^{k} \left[ {k \atop i} \right] \zeta(i+1) - \sum_{i=0}^{k} \left[ {k \atop i} \right] \bigg{)} \bigg{)}. \end{align}.$$

Now, we know that $\zeta(1) = \infty$ and that $\left[ {k \atop 0} \right] = 0$. In order to evaluate the sum above, I thus wonder:

Question: what is $\lim_{t \to 0} \left[ {k \atop t} \right] \zeta(t+1) $ ?

Max Muller
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  • @Conrad yes good point, I'm also not entirely sure what $\left[ {k \atop t} \right] $ is for non-integer $t$. It should be some sort of analytic continuation of the unsigned Stirling numbers of the first kind. What you wrote is a binomial interpretation, no? – Max Muller Nov 25 '21 at 21:10
  • yers - didn't really read carefully what you wrote so deleted the comment; not sure about rearranging sums though as one usually needs absolute convergence for that; it may work sure, but one needs proof – Conrad Nov 25 '21 at 21:15
  • @Conrad you're right, that requires proof too. Even if rearrangement turns out not to be justified, I'm still interested in the answer to the question – Max Muller Nov 25 '21 at 21:18
  • Note that in the first sum, the innermost sum has $0$ for $i=0$ so in principle that should allow you to disregard that and ignore the $\zeta(1)=\infty$ – Conrad Nov 25 '21 at 21:20
  • If swapping sums yields a series that doesn’t converge, can you really do it? – Thomas Andrews Nov 25 '21 at 21:20
  • @Thomas - in practice, if a double sum converges conditionally, there is usually a way to swap, but it may require cleverness or some notion of summability instead (and of course proof that you are correct) - here though i think that the divergent term can be ignored as noted above – Conrad Nov 25 '21 at 21:23
  • @ThomasAndrews I'm not entirely sure how you can infer swapping sums yields a series that doesn't converge in this case - could you elaborate? – Max Muller Nov 25 '21 at 21:26
  • @Conrad an overview of analytic continuations of the stirling numbers of the first kind can be found here: https://www.tandfonline.com/doi/abs/10.1080/10236190902780190 – Max Muller Nov 25 '21 at 21:29

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