Proof that alterntating zeta series converges for $\mathrm{Re}(s)\ge0$ I found this proof:

I do not understand the motivation in going from the first to the second line. What convergence test, if any, have they actually used? Could someone just give more details on how they attempted this proof.
 A: Consider the definition of the derivative: $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}.$$ In the summation, you have something like $$\frac{f(n+1)-f(n)}{1}.$$So all that is happening is that they are making an estimation of the derivative:
$$ \left|\frac{f(2n-1)-f(2n)}{1}\right|\leq\max_{x\in[2n-1,2n]}\left\{\left|\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\right|\right\}\leq\left|\frac{s}{(2n-1)^{s+1}}\right|.$$
For a proof of this inequality, see this page.
Once they have this, the resulting sum is absolutely convergent since $s$ is fixed (in particular, $|s|\leq M$) and $|(2n-1)^{s+1}|=(2n-1)^{\mathrm{Re}(s)+1}$. In particular, $$\sum_{n=1}^\infty\frac{|s|}{(2n-1)^{1+\mathrm{Re}(s)}}\leq\sum_{n=1}^\infty\frac{M}{n^{1+\mathrm{Re}(s)}}.$$The latter sum can be shown to converge using the integral test.
A: I think the proof you posted is using the MVT for the function $\;f(z):=\cfrac1{z^s}\;$ on the interval $\;[2n-1,\,2n]\;$ , since then
$$\frac{f(2n)-f(2n-1)}{2n-(2n-1)}=f'(c)\;,\;\;c\in (2n-1,\,2n)$$
But
$$\frac{f(2n)-f(2n-1)}{2n-(2n-1)}=f(2n)-f(2n-1)=\frac1{(2n)^s}-\frac1{(2n-1)^s}$$
and
$$f'(c)=-\frac s{c^{s+1}}$$
and taking absolute values we get at once
$$|f'(c)|=\left|\frac s{c^{s+1}}\right|\le\frac{|s|}{|2n-1|^{s+1}}\,,\,\,\text{since}\;\;c>2n-1...$$
and the series of the last rightmost expression converges say because $\;\text{Re}\,(s+1)>1\;$ ...
