How to show $\zeta (1+\frac{1}{n})\sim n$ How to show $\zeta (1+\frac{1}{n})\sim n$ as $n\rightarrow \infty$ where $\zeta$ is the Riemann zeta function.
And can we say $\lceil \zeta (1+\frac{1}{n}) \rceil=n$ for any positive integer $n\geq 1$. How to prove it?
 A: HINT:
Note that we have
$$\int_2^\infty \frac{1}{x^{1+1/n}}\,dx\le \sum_{k=2}^\infty\frac{1}{k^{1+1/n}}\le \int_1^\infty \frac{1}{x^{1+1/n}}\,dx \tag 1$$

SPOILER ALERT:  Scroll over the highlighted area to reveal the soluiton

Evaluating the right-hand side of $(1)$, we see that $$\int_1^\infty \frac{1}{x^{1+1/n}}\,dx=n \tag 2$$ while evaluating the left-hand side of $(1)$, we see that $$\int_2^\infty \frac{1}{x^{1+1/n}}\,dx=n2^{-1/n} \tag 3$$Then, note that $$n2^{-1/n}\ge n-\log(2) \tag 4$$Using $(2)-(4)$ in $(1)$, we find that $$n+(1-\log(2)) \le \sum_{k=1}^\infty\frac{1}{k^{1+1/n}}\le n+1$$from which we conclude that $$\sum_{k=1}^\infty\frac{1}{k^{1+1/n}}\sim n$$

A: In this answer, it is shown that near $s=1$
$$
\zeta(s)=\frac1{s-1}+\gamma+O(s-1)
$$
Therefore,
$$
\zeta\left(1+\frac1n\right)=n+\gamma+O\left(\frac1n\right)
$$
A: According to this page
$$\zeta(s)=\frac{1}{s-1}+\sum_{n=0}^\infty \frac{(-1)^n}{n!} \gamma_n \; (s-1)^n$$
the terms $γ_n$ being Stieltjes constants. So, if $s=1+\frac 1n$ $$\zeta\big(1+\frac 1n\big)=n+\sum_{i=0}^\infty \frac{(-1)^i}{i!} \gamma_i \; \frac 1{n^i}=n+\gamma+\sum_{i=1}^\infty \frac{(-1)^i}{i!}  \; \frac{\gamma_i } {n^i}$$
A: Is $\displaystyle \left.\frac d {ds} \, \frac 1 {\zeta(s)}\right|_{s=1}$ equal to $1$?  If so, then $$ 1 = \lim_{s=1} \frac{\dfrac 1 {\zeta(s)} - \dfrac 1 {\zeta(1)}}{s-1} = \lim_{s=1} \frac{\left(\dfrac 1 {\zeta(s)}\right)}{s-1}, $$
so
$$
\zeta(s) \sim \frac 1 {s-1} \text{ as }s\to 1.
$$
Apply this when $s=1 + \dfrac 1 n$.
