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?
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 $$\sum_{k=1}^\infty\frac{1}{k^{1+1/n}}\sim n$$
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 $s=1 + \dfrac 1 n$.