How to find $a_r(x)$ such that: $n^{x-1 }(\zeta(x)-\sum_{k=1}^n \frac{1}{k^x})= \frac{1}{x-1} +\sum_{r\ge 1} \frac{a_r(x)}{n^r}$

Question

I found this question : For $x>1$ find $$\lim\limits_{n \to \infty} n\left(n^{x-1 }\left(\zeta(x)-\sum\limits_{k=1}^n \frac{1}{k^x}\right)-\frac{1}{x-1}\right)$$

Using Stolz-Cesàro theorem:

Stolz-Cesàro theorem case $\frac{0}{0}$:- If $b_n$ is a strictly decreasing sequence and $\lim \limits_{n \to \infty} b_n = \lim \limits_{n \to \infty} a_n = 0$, and if $\lim \limits_{n \to \infty} \frac{a_{n+1}-a_n}{b_{n+1}- b_n}= l \in \overline{\mathbb{R}}$, then $\lim \limits_{n \to \infty} \frac{a_n }{b_n}=l$.

It is easy to prove that: $$\lim\limits_{n \to \infty} n^{x-1 } (\zeta(x)-\sum\limits_{k=1}^n \frac{1}{k^x})=\frac{1}{x-1}$$

However it is not easy to prove that $n^{x-1 } \left(\zeta(x)-\sum\limits_{k=1}^n \frac{1}{k^x}\right)$ is a monotone sequence and even if we proved that applying Stolz-Cesàro theorem would be very difficult.

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I also want to ask for a generalization: Find a closed form for $a_r(x)$ such that:

$$n^{x-1 } \left(\zeta(x)-\sum\limits_{k=1}^n \frac{1}{k^x}\right)= \frac{1}{x-1} +\sum\limits_{r\ge 1} \frac{a_r(x)}{n^r}$$

Answer 1

Too long for a comment

This is an ideal problem for the Euler-Maclaurin summation formula application $$\sum_{k=1}^n\frac1{k^x}=\sum_{k=1}^{n-1}\frac1{k^x}+\frac1{n^x}=\zeta(x)-\sum_{k=n}^\infty\frac1{k^x}+\frac1{n^x}$$ Using the formula $$\sum_{k=n}^\infty\frac1{k^x}\sim\int_n^\infty\frac{dk}{k^x}+\frac1{2n^x}+\sum_{l=1}^\infty\frac{B_{2l}}{(2l)!}\frac{x(x+1)...(x+2l-2)}{n^{x+2l-1}}$$ $$\sim\frac1{(x-1)n^{x-1}}+\frac1{2n^x}+\sum_{l=1}^\infty\frac{B_{2l}}{(2l)!}\frac{\Gamma(x+2l-1)}{\Gamma(x)}\frac1{n^{x+2l-1}}$$ and $$\left(n^{x-1 }\left(\zeta(x)-\sum\limits_{k=1}^n \frac{1}{k^x}\right)-\frac{1}{x-1}\right)\sim-\frac1{2n}+\sum_{l=1}^\infty\frac{B_{2l}}{(2l)!}\frac{\Gamma(x+2l-1)}{\Gamma(x)}\frac1{n^{2l}}$$ The limit follows.