I have been working my way through this Arxiv paper concerning the analytic continuation of the zeta function. I don't understand the first equality in equation (19), page 6.
In equation (11), the paper gives the definition
$$\eta(s):=\sum _{n=1}^{\infty } \frac{(-1)^{n+1}}{n^s}$$
Equation (14) summarises results so far as
$$\zeta (s)=\frac{1}{1-2^{1-s}}\sum _{n=1}^{\infty } \frac{(-1)^{n+1}}{n^s}=\frac{\eta(s)}{1-2^{1-s}}=\frac{1}{s-1}\sum _{n=1}^{\infty} \left(\frac{n}{(n+1)^s}-\frac{n-s}{n^s}\right)$$
which is fine. But equation (19) then asserts that
$$\underset{s\to 1}{\text{lim}}\zeta(s)=\underset{s\to 1}{\text{lim}}\frac{s-1}{1-2^{1-s}}\sum_{n=1}^{\infty } (-1)^{n+1}n^{-s}$$
Doubtless I'm being unobservant / stupid - but where does the factor $s-1$ come in? In equation (14) we have $\zeta (s)=\frac{1}{1-2^{1-s}}\eta(s)$ - and now in equation (19) we have $\underset{s\to 1}{\text{lim}}\zeta(s)=\underset{s\to 1}{\text{lim}}\frac{s-1}{1-2^{1-s}}\eta(s)$...
How and why does the extra factor $s-1$ appear? Presumably it's something to do with the limit-taking process, but I don't understand.
