How does $\zeta(1 - s)$ become $(-1/s + \cdots)$? Why is
$$\zeta(1 - s) = -\frac{1}{s} + \cdots$$
for small negative values of $s$?
A detailed explanation would be appreciated.
 A: For $s < 0$
$$ \begin{eqnarray}
    \zeta(1-s) &=& \sum_{n=1}^\infty n^{s-1} = \int_1^\infty x^{s-1} \mathrm{d}x + \sum_{n=1}^\infty \left( n^{s-1} - \int_{n}^{n+1} x^{s-1} \mathrm{d}x \right) \\ &=& -\frac{1}{s} + \sum_{n=1}^\infty \left( n^{s-1} - \frac{(n+1)^s - n^s}{s} \right)
\end{eqnarray}
$$
Now, observe that the following limit is finite
$$
    \lim_{s \to 0^-} \zeta(1-s) + \frac{1}{s} = \sum_{n=1}^\infty  \left( \frac{1}{n} - \log\frac{n+1}{n} \right) = \gamma \tag{1}
$$
where $\gamma$ is the Euler-Mascheroni constant. The sum on the right-hand-side of $(1)$ converges , since for large $n$
$$
   \frac{1}{n} - \log\left( 1+\frac{1}{n}\right) = \frac{1}{2 n^2} + \mathcal{o}\left(\frac{1}{n^2}\right)
$$
Alternatively:
$$
  \sum_{n=1}^\infty  \left( \frac{1}{n} - \log\frac{n+1}{n} \right) = \lim_{m \to \infty} \sum_{n=1}^m\left( \frac{1}{n} - \log\frac{n+1}{n} \right) = \lim_{m \to \infty} \left( \sum_{n=1}^m \frac{1}{n} - \log(m+1) \right) = \gamma
$$
Hence, this establishes that, for small negative $s$:
$$
    \zeta(1-s) = -\frac{1}{s} + \gamma + \mathcal{o}(1)
$$
