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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?

esege
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    the simpler is $\zeta(s) = s\int_1^\infty \lfloor x \rfloor x^{-s-1}dx = \frac{s}{s-1} + \mathcal{O}(s)$ for $Re(s) \ge 1$, or $\eta(s) = (1-2^{1-s}) \zeta(s) = \mathcal{O}(1)$ for $Re(s) \ge 1$ – reuns Feb 20 '16 at 01:10
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    ok, Back to a pet peeve: "Is WHATEVER true for any positive integer $n$?" can be construed to mean "Is there any positive integer $n$ for which WHATEVER holds?". That makes "any" in this context almost synonymous with "some", but I don't think that is what was meant. Just changing it to "every" would fully disambiguate it. English-speaking mathematicians use the word "any" too much. $\qquad$ – Michael Hardy Feb 20 '16 at 01:10
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    Even better $\zeta(1+1/n) \sim n+\gamma_0$ where $\gamma_0$ is the Eller constant. – Mhenni Benghorbal Feb 20 '16 at 01:42
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    @MhenniBenghorbal: that is what my answer says. – robjohn Feb 20 '16 at 01:47
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    https://en.wikipedia.org/wiki/Riemann_zeta_function#Laurent_series – Will Jagy Feb 20 '16 at 01:54
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    @robjohn: I saw your answer after I posted my comment! I solved similar problems before! – Mhenni Benghorbal Feb 20 '16 at 01:57
  • @robjohn: By the way sometimes I post a comment or an answer before I refresh the page that is why you do not see some new comments or answers! – Mhenni Benghorbal Feb 20 '16 at 02:10
  • @robjon Rob. Actually, I didn't. Rather, I just bounded the sum. - Mark – Mark Viola Feb 20 '16 at 04:28

4 Answers4

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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 $$\sum_{k=1}^\infty\frac{1}{k^{1+1/n}}\sim n$$

Mark Viola
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In this answer, it is shown that near $s=1$ $$ \zeta(s)=\frac1{s-1}+\gamma+O(s-1) $$ Therefore, $$ \zeta\left(1+\frac1n\right)=n+\gamma+O\left(\frac1n\right) $$

Community
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robjohn
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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 $s=1+\frac 1n$ $$\zeta\big(1+\frac 1n\big)=n+\sum_{i=0}^\infty \frac{(-1)^i}{i!} \gamma_i \; \frac 1{n^i}=n+\gamma+\sum_{i=1}^\infty \frac{(-1)^i}{i!} \; \frac{\gamma_i } {n^i}$$

Claude Leibovici
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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 $s=1 + \dfrac 1 n$.

Michael Hardy
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