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How fast does $\zeta(k)$ grow where $k\in\Bbb N$?

Is it polynomial in $k$?

($\zeta$ is the Riemann zeta function.)

Caleb Stanford
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Turbo
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    If $\zeta$ denotes the Riemann $\zeta$-function, then $\zeta(k) \to 1$ as $k \to \infty$. – Daniel Fischer Nov 29 '16 at 18:56
  • u might find this useful:http://math.stackexchange.com/questions/1630809/calculate-or-estimate-sx-sum-k-1-infty-frac-zetakxk/1630884#1630884 – tired Nov 29 '16 at 18:58
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    the limit $k\rightarrow -\infty$ might be explored via the functional equation – tired Nov 29 '16 at 19:00
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    for $k > 1$ : $ \frac{3^{1-k}}{k-1}=\int_3^\infty x^{-k}dx <\sum_{n=3}^\infty n^{-k} < \int_2^\infty x^{-k}dx=\frac{2^{1-k}}{k-1} $ so $$1+2^{-k}+\frac{3^{1-k}}{k-1} < \zeta(k) < 1+2^{-k}+\frac{2^{1-k}}{k-1}$$ – reuns Feb 06 '17 at 11:53

1 Answers1

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As Daniel Fischer pointed out, $\zeta(k) \to 1$ as $k \to \infty$. This is expected: recall the definition of the $\zeta$ function: $$ \zeta(k) = 1 + \frac{1}{2^k} + \frac{1}{3^k} + \frac{1}{4^k} + \cdots \quad (\text{for } k > 1) $$ Since the series converges, and each term other than the first is monotonically decreasing to $0$ as $k \to \infty$, every term drops out in the limit except the initial term $1$. So the entire series approaches $1$ as $k \to \infty$.${}^1$

You can get a better approximation by taking more terms. In particular, $\frac{1}{2^k}$ decreases to $0$ much more slowly (although still very quickly) than $\frac{1}{3^k}$, so $$ \zeta(k) -1 \sim \frac{1}{2^k} \text{ as } k \to \infty, $$ which should be a pretty good approximation.

${}^1$More generally: if $f(k) = \sum_{i = 1}^\infty f_i(k)$, the series converges for all $k$, and $f_i(k)$ is positive and monotonically decreasing in $\boldsymbol{k}$, then $\lim_{k \to \infty} f(k) = \sum_{i = 1}^\infty \lim_{k \to \infty} f_i(k)$. This is a consequence of the Monotone convergence theorem; alternatively, it's not too hard to prove directly.

Caleb Stanford
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    by comparison with $\int_3^\infty x^{-k} dx$ : $\ \ 1+2^{-k}+\frac{3^{1-k}}{k-1} < \zeta(k) < 1+2^{-k}+\frac{2^{1-k}}{k-1}$ – reuns Feb 06 '17 at 11:55
  • @6005 I am looking for tight upper and lower bounds for $\frac1{\zeta(k)}$. Is it $1-\frac1{2^{k-1}}<\frac1{\zeta(k)}<1-\frac1{2^k}$? – Turbo May 11 '22 at 21:57
  • @Turbo For $k > 1$, or $0 < \Re k < 1$, or general $k$? (You might want to ask a new question) – Caleb Stanford May 12 '22 at 04:30
  • @6005 General integer $k\geq1$ – Turbo May 12 '22 at 08:46
  • @Turbo An upper bound cannot be derived directly from this answer, it only gives a trivial lower bound $\zeta(k) > 1 + \frac{1}{2^k}$. These notes appear to have an upper bound (see Thm 6.25). If that doesn't fully help you, please ask a new question. – Caleb Stanford May 12 '22 at 16:53
  • @6005 https://math.stackexchange.com/questions/4449074/what-is-a-good-upper-bound-for-zetak – Turbo May 12 '22 at 17:21