How to prove $\lim_{s \rightarrow \infty} \zeta(s) = 1$?

Question

$\lim_{s \rightarrow \infty} \zeta(s) = 1$

I have seen a proof using the fact $1 \leq \zeta(s) \leq \frac{1}{1-2^{1-s}}$ but this relies on proving the inequality first which is quite cumbersome.

I was looking for a proof which is short but easy to understand.

Answer 1

Despite your aversion(?) to integrals, I think they give the quickest and easiest argument. The function $n \mapsto n^{-s}$ is decreasing in $n$ (for $s > 1$), so $$ 1 \le \zeta(s) = \sum_{n=1}^\infty \frac1{n^s} = 1 + \sum_{n=2}^\infty \frac1{n^s} \le 1 + \int_1^\infty \frac{1}{x^s}\,ds = 1 + \frac{1}{s-1}. $$ To see why the inequality between the series and the integral is valid, draw a picture! Finish off by letting $s \to \infty$ using the squeeze theorem. (This argument assumes that $s$ is real.)

Answer 2

If we know that $\zeta$ is uniformly convergent then $$\lim_s \zeta (s)=\lim_{s}\sum_{n=1}^\infty \frac1{n^s}=\sum_{n=1}^\infty \lim_s \frac1{n^s}=1+0+0+\ldots =1.$$

Answer 3

EDIT: Upon reading the initial comments I see this is probably the proof the poster already saw but thought was too cumbersome. Oh well here it is:

Let's modify the classic proof that the harmonic series diverges. First we'll note that $\zeta(s)>1$ for large $s$ trivially. Next, note that for $s>1$ we have the following:

$$1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\frac{1}{5^s}+\dots < 1+\frac{1}{2^s}+\frac{1}{2^s}+\frac{1}{4^s}+\frac{1}{4^s} \dots$$

Where we replace $\frac{1}{n^s}$ by $\frac{1}{2^{ks}}$ where $2^k$ is the largest power of 2 less than or equal to $n$. Now combining terms on the right hand side we get:

$$1 +\frac{2}{2^s} + \frac{4}{4^s} +\frac{8}{8^s}+\dots = \sum_{i=0}^\infty \big(\frac{1}{2^{s-1}}\big)^i$$

Finally summing this geometric series gives:

$$\zeta(s) \le \frac{1}{1-\frac{1}{2^{s-1}}} = 1 + \frac{1}{2^{s-1}-1}$$

Which clearly goes to $1$ as $s$ tends to infinity.

Answer 4

Here is a generic proof, in the sense that it generalizes to many other kinds of series:

First, note that any partial sum of the series converges: $$ \lim_{\zeta\to\infty} \sum_{k=1}^n \frac{1}{k^\zeta}=1. $$ This is trivial, as the first term is constant $1$ and the others go to zero.

Second, note that given any fixed $\zeta>1$ and $\varepsilon>0$, you can make the tail small, just by picking $n$ sufficiently large: $$ \sum_{k=n+1}^\infty\frac{1}{k^\zeta}<\varepsilon. $$ This is because the series is convergent. Further, note that each term is a decreasing function of $\zeta$, hence so is the sum of the tail, so the above inequality holds for $n$ and $\zeta$ sufficiently large.

Now combine the two parts to get $$\sum_{k=1}^\infty \frac{1}{k^\zeta}<1+2\varepsilon$$ for $\zeta$ sufficiently large. Since the sum is also $>1$ (trivially), we're done.

(This is basically just a reworking of the dominated convergence theorem for sums.)

Answer 5

The following is a very simple argument. Is it flawed?

1. The magnitude of each summand is $|n^{-s}|=|n^{-\sigma-it}|=n^{-\sigma}$.

2. As $s\rightarrow\infty$ so does $\sigma\rightarrow\infty$

3. The magnitude of all the terms in the series $\zeta(s)=\sum1/n^s$ therefore $\rightarrow 0$ except the very first which remains 1.

Therefore $\zeta(s)\rightarrow1$ as $s\rightarrow\infty$.

Answer 6

Even though it's nice to reprove the dominated convergence theorem, we don't need to and can simply apply it: $$ \lim_{s\to \infty} \zeta(s) = \lim_{s\to \infty} \sum_{n=1}^\infty \frac1{n^s} = \sum_{n=1}^\infty \lim_{s\to \infty} \frac1{n^s} = 1 + \sum_{n=2}^\infty 0=1. $$ You are allowed to interchange the limits because the series is dominated by $\zeta(2)<\infty$ for example for $s>2$.