<s>[INST] Limit of Zeta function I'm looking for a reference for (or an elementary proof of)
$$ \lim_{s \rightarrow 1} \left( \zeta(s) - \frac{1}{s-1} \right) = \gamma$$
Thanks for your help.
 [/INST] Elementary proof:
For $s>1$, write
$$\zeta(s) - \frac{1}{s-1} = \sum_{n=1}^{\infty} \left( \frac{1}{n^s} - \int_{n}^{n+1} \frac{dx}{x^s} \right).$$
I will show below that we can interchange limits and summation. Assuming this, we have
$$\lim_{s \to 1^{+}} \left(  \zeta(s) - \frac{1}{s-1} \right) = \sum_{n=1}^{\infty} \left( \frac{1}{n} - \int_{n}^{n+1} \frac{dx}{x} \right).$$
(It is obvious that we can interchange limits with integration by just explicitly computing the integral.)
The sum of the first $N$ terms of the above is
$$\sum_{n=1}^{N} \frac{1}{n} - \int_1^{N+1} \frac{dx}{x} = \sum_{n=1}^{N} \frac{1}{n} - \log(N+1).$$
So
$$\lim_{s \to 1^{+}} \left(  \zeta(s) - \frac{1}{s-1} \right) = \lim_{N \to \infty} \left( \sum_{n=1}^{N} \frac{1}{n} - \log(N+1) \right)$$
and the right hand side is, by definition $\gamma$.

We now justify interchanging limits. Set 
$$a_n(s) = \frac{1}{n^s} - \int_{n}^{n+1} \frac{dx}{x^s}=\int_n^{n+1} \left( \frac{1}{n^s} - \frac{1}{x^s} \right) dx.$$ 
We are going to use the Weierstrass $M$-test, which means that we must find a convergent series $\sum M_n$ such that, for all $s$, we have $|a_n(s)| < M_n$. Actually, it is enough to do for $s$ in the interval $[1,2]$, so I will assume this from now on. I'm sure there are many ways to come up with such a boundl this is the one I found first.
We have
$$a_n(s) < \int_n^{n+1} \left( \frac{1}{n^s} - \frac{1}{(n+1)^s} \right) dx = \frac{1}{n^s} - \frac{1}{(n+1)^s}=\frac{1}{n^s} \left( 1- \frac{1}{(1+1/n)^s} \right)$$
$$ < \frac{1}{n} \left( 1- \frac{1}{(1+1/n)^2} \right) < \frac{1}{n} \cdot \frac{2}{n}.$$
At the line break, we used that $1/n^s$ is maximized at the left endpoint of $[1,2]$, and $1-1/(1+1/n)^s$ is maximized at the right endpoint.
So we take $M_n = 2/n^2$ and we are done.
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