# Limit of a Function Involving Hurwitz Zeta Function

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.

• 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 Feb 29 at 18:26
• @Conrad yes, it is. Feb 29 at 18:36

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 $$p$$.