I am trying to prove the following limit of a function involving the Hurwitz Zeta function:

$$ \lim_{N \to \infty} \frac{\zeta(-d, 1 + N) - \zeta(-d, 1 + p N)}{N^{1 + d}} = \frac{-1 + p^{1 + d}}{1 + d} $$

where $d$ is a positive integer, and $p$ is a real number such that $0 < p < 1$.

I have tried to prove this limit directly, but I am not sure how to handle the Hurwitz Zeta function in this context.

Any help or insights on how to approach this problem would be greatly appreciated.

  • $\begingroup$ to be clear I presume that here you take $\zeta(s,a)=\sum_{n \ge 0}(n+a)^{-s}, \Re s >1, a >0$ where $\zeta(s,1)$ is the usual RZ $\endgroup$
    – Conrad
    Feb 29 at 18:26
  • $\begingroup$ @Conrad yes, it is. $\endgroup$ Feb 29 at 18:36

1 Answer 1


We can use the identity $$\zeta(-d, a)= -\frac{B_{d+1}(a)}{d+1}, \quad (a>0, \ d \in \mathbb{N}), $$ where $B_{d}(a)$ is the $dth$ Bernoulli polynomial. This identity is typically obtained from the Hankel contour integral representation of the Hurwitz zeta function.

As $ a \to +\infty$, $B_{d+1}(a)= \sum_{k=0}^{d+1} \binom{d+1}{k}B_{d+1-k} \, a^{k}$ is asymptotic to $a^{d+1}$.

Therefore, $\zeta(-d,a) $ is asymptotic to $- \frac{a^{d+1}}{d+1}$ as $a \to + \infty$, and $$ \begin{align} \lim_{N \to \infty} \frac{\zeta(-d, 1 + N) - \zeta(-d, 1 + p N)}{N^{d+1}} &= \lim_{N \to \infty}\frac{-(1+N)^{d+1}+(1+pN)^{d+1}}{N^{d+1}(d+1)} \\ &= \lim_{N \to \infty} \frac{-N^{d+1} \left(\frac{1}{N}+1 \right)^{d+1} + N^{d+1} \left(\frac{1}{N}+p \right)^{d+1}}{N^{d+1}(d+1)} \\ &= \lim_{N \to \infty} \frac{-\left(\frac{1}{N}+1 \right)^{d+1} + \left(\frac{1}{N}+p \right)^{d+1}}{d+1} \\ &= \frac{-1+p^{d+1}}{d+1}. \end{align}$$

The limit holds for all positive values of $p$.


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