How to prove $\lim_{s \rightarrow \infty} \zeta(s) = 1$? $\lim_{s \rightarrow \infty} \zeta(s) = 1$
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.
 A: 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.)
A: 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.
A: 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.)
A: If we know that $\zeta$ is uniformly convergent then
$$\lim_s \zeta (s)=\lim_{s}\sum_{n=1}^\infty \frac1{n^s}=\sum_{n=1}^\infty \lim_s \frac1{n^s}=1+0+0+\ldots =1.$$
A: 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 $s>2$.
A: The following is a very simple argument. Is it flawed?

*

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


*As $s\rightarrow\infty$ so does $\sigma\rightarrow\infty$


*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 $s\rightarrow\infty$.
