This question loosely builds this one.
Equate the following two infinite series for $\zeta(s)$:
$$\displaystyle \zeta(s) = \frac{1}{4\,(s-1)} \left(1+s+\sum _{n=1}^{\infty } \left( {\frac {s+1+3\,n}{(n+1)^{s}}} + \frac{3\,s-2-2\,n}{n^s}-{\frac {n-1}{\left(n-1 \right) ^{s}}}\right) \right) \quad -1<\Re(s)<1$$
and
$$\displaystyle \zeta(s) = \frac{1}{4\,(s-1)} \left(1+\sum _{n=1}^{\infty } \left( {\frac {2n+1}{(n+1)^{s}}} + \frac{4\,s-3}{n^s}-{\frac {2\,(n-1)}{\left(n-1 \right) ^{s}}}\right) \right) \quad -1<\Re(s)<1$$
Remove the factor $\frac{1}{4\,(s-1)}$ from both sides and then the difference between the remaining parts must be $-s$, so that:
$$\displaystyle f(s) = \sum _{n=1}^{\infty } \left( {\frac {-s-n}{(n+1)^{s}}} + \frac{s-1+2 \,n}{n^s}-{\frac {n-1}{\left(n-1 \right) ^{s}}}\right) =s \qquad -1<\Re(s)<1$$
Obviously quite a complex way to express something as simple as $f(s)=s$ (in this restricted domain), however I am keen to understand whether an easier explanation exists for this phenomenon, rather than having to take the difference between the two infinite series.
Thanks.
====================== UPDATE =====================
I think I found it :-)
For $n=1$ the middle term becomes $s+1$. For $n=2$ the last term becomes $-1$ and together that gives $s$. All the other terms 'telescope' each other out up till a finite $N$ where we are left with:
$$\lim_{N \to +\infty} \left(\frac{N}{N^{s}}-\frac{N+s}{(N+1)^s} \right) =0 \qquad \Re(s)>-1$$