Growth of Riemann zeta function $\zeta(k)$

Question

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

Is it polynomial in $k$?

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

Answer 1

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.

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${}^1$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.