<s>[INST] Zeta Infinite Summation $\sum_{n=1}^{\infty}(-1)^{n-1}\,\zeta(s+n)$ Let $Re\{s\}\gt0$ :
$$
\sum_{n=1}^{\infty} \frac{n^{-s}}{n+1} = \sum_{n=1}^{\infty} \left( \frac{n}{n+1}\right) n^{-(s+1)} = \sum_{n=1}^{\infty} \left( 1 - \frac{1}{n+1}\right) n^{-(s+1)} = \zeta(s+1) - \sum_{n=1}^{\infty} \frac{n^{-(s+1)}}{n+1} = \\[6mm]
\zeta(s+1) - \zeta(s+2) + \sum_{n=1}^{\infty} \frac{n^{-(s+2)}}{n+1} = \zeta(s+1) - \zeta(s+2) + \zeta(s+3) - \sum_{n=1}^{\infty} \frac{n^{-(s+3)}}{n+1} = \text{...} \Rightarrow \\[6mm]
\sum_{n=1}^{N} (-1)^{n-1} \zeta(s+n) = \sum_{n=1}^{\infty} \frac{n^{-s}}{n+1} - (-1)^{N}\sum_{n=1}^{\infty} \frac{n^{-(s+N)}}{n+1} \Rightarrow \\[6mm]
\boxed{ \quad \sum_{n=1}^{\infty} (-1)^{n-1} \zeta(s+n) = \sum_{n=1}^{\infty} \frac{n^{-s}}{n+1} - \lim_{N\rightarrow\infty}\left[(-1)^{N}\sum_{n=1}^{\infty} \frac{n^{-(s+N)}}{n+1}\right] \quad } \\[6mm]
$$

Does the limit exist? and What does it equal?
  $$ L = \lim_{N\rightarrow\infty}\left[(-1)^{N}\sum_{n=1}^{\infty} \frac{n^{-(s+N)}}{n+1}\right] \qquad\qquad\colon\space Re\{s\} \gt 0 \tag{1}$$
  $$\sum_{n=1}^{\infty} (-1)^{n-1} \zeta(s+n) = \sum_{n=1}^{\infty} \frac{n^{-s}}{n+1} - L \qquad\colon\space Re\{s\} \gt 0 \tag{2}$$


Without the outer sign, the limit is:
$$
\small \sum_{n=2}^{\infty}\frac{n^{-(s+N)}}{n^2}\lt\sum_{n=2}^{\infty}\frac{n^{-(s+N)}}{n+1}\lt\sum_{n=2}^{\infty}\frac{n^{-(s+N)}}{n} \Rightarrow \zeta(s+N+2)-1 \lt \sum_{n=2}^{\infty}\frac{n^{-(s+N)}}{n+1} \lt \zeta(s+N+1)-1 \\
\small \text{Let}\space\left\{N\rightarrow\infty\right\}\space\text{and use the limit}\space\left\{\lim_{z\rightarrow\infty}\zeta(z)=1\right\} \Rightarrow \lim_{N\rightarrow\infty}\sum_{n=2}^{\infty}\frac{n^{-(s+N)}}{n+1}=0 \Rightarrow \color{red}{\lim_{N\rightarrow\infty}\sum_{n=1}^{\infty}\frac{n^{-(s+N)}}{n+1}=\frac{1}{2}} \\
$$
NB: appreciating your explanations on a similar previous question. Many Thanks.

 
conclusion:
As of the correct answer(s):
$$
\sum_{n=1}^{\infty}\frac{n^{-s}}{n+1} - \sum_{n=1}^{\infty}(-1)^{n-1}\,\zeta(s+n) = \sum_{n=1}^{\infty}\frac{n^{-s}}{n+1} - \sum_{n=1}^{\infty}\left[\color{red}{\zeta(s+2n-1)-\zeta(s+2n)}\right] \\[6mm]
\quad = \lim_{N\rightarrow\infty}\sum_{n=1}^{\infty}\frac{n^{-(s+N)}}{n+1} = \color{red}{\frac{1}{2}} \quad\colon\space Re\{s\}\ge0 \quad\{\small\text{holds for s=0 too}\normalsize\} \\[6mm]
$$
 [/INST] Let
$S(N)
=\sum_{n=1}^{N} \frac{n^{-s}}{n+1}
=\sum_{n=1}^{N} \frac{1}{n^s(n+1)}
$.
If $Re(s)> 0$,
$\lim_{N \to \infty} S(N)$
exists by comparison with
$\sum_{n=1}^{N} \frac{1}{n^{1+Re(s)}}
$.
Then,
since
$\frac1{1+x}
=\sum_{k=0}^{2m} (-1)^k x^k
-\frac{x^{2m+1}}{1+x}
$
$\begin{array}\\
S(N)
&=\sum_{n=1}^{N} \frac{1}{n^s(n+1)}\\
&=\sum_{n=1}^{N} \frac{1}{n^{s+1}(1+1/n)}\\
&=\sum_{n=1}^{N} \frac{1}{n^{s+1}}(\sum_{k=0}^{2m} (-1)^k n^{-k}
-\frac{n^{-2m-1}}{1+1/n})\\
&=\sum_{n=1}^{N} \frac{1}{n^{s+1}}\sum_{k=0}^{2m} (-1)^k n^{-k}
-\sum_{n=1}^{N} \frac{1}{n^{s+1}}\frac{n^{-2m-1}}{1+1/n}\\
&=\sum_{n=1}^{N} \sum_{k=0}^{2m} (-1)^k \frac1{n^{k+s+1}}
-\sum_{n=1}^{N} \frac{1}{(1+1/n)n^{s+2m+2}}\\
&= \sum_{k=0}^{2m}\sum_{n=1}^{N} (-1)^k \frac1{n^{k+s+1}}
-\sum_{n=1}^{N} \frac{1}{(n+1)n^{s+2m+1}}\\
&= \sum_{k=0}^{2m}(-1)^k\sum_{n=1}^{N}  \frac1{n^{k+s+1}}
-\sum_{n=1}^{N} \frac{1}{(n+1)n^{s+2m+1}}\\
&= S_1(N)-S_2(N)\\
\end{array}
$
As $N \to \infty$,
$S_1(N)
= \sum_{k=0}^{2m}(-1)^k\sum_{n=1}^{N}  \frac1{n^{k+s+1}}
\to \sum_{k=0}^{2m}(-1)^k \zeta(k+s+1)
$
and
$\begin{array}\\
S_2(N)
&=\sum_{n=1}^{N} \frac{1}{(n+1)n^{s+2m+1}}\\
&\gt \frac12\\
\text{and}\\
S_2(N)
&=\frac12+\sum_{n=2}^{N} \frac{1}{(n+1)n^{s+2m+1}}\\
&\le\frac12+\sum_{n=2}^{N} \frac{1}{(n+1)2^{2m}n^{s+1}}\\
&=\frac12+\frac1{4^m}\sum_{n=2}^{N} \frac{1}{(n+1)n^{s+1}}\\
&\lt\frac12+\frac1{4^m}\sum_{n=2}^{N} \frac{1}{(n+1)n}\\
&=\frac12+\frac1{4^m}\sum_{n=2}^{N} (\frac1{n}-\frac{1}{n+1})\\
&<\frac12+\frac1{2\cdot 4^m}\\
\end{array}
$
Therefore
$\lim_{m \to \infty} S_2(N)
= \frac12
$,
so that
$\lim_{m \to \infty, N \to \infty} S(N)
=-\frac12+\sum_{k=0}^{2m}(-1)^k \zeta(k+s+1)
$.
Note that,
if we write
$\sum_{k=0}^{2m}(-1)^k \zeta(k+s+1)
=\sum_{k=0}^{m-1}(\zeta(2k+s+1)-\zeta(2k+s+2))
+\zeta(2m+s+1)
$,
the sum converges properly.
If we do this grouping
earlier on
(even and odd terms together
with opposite signs),
this all becomes rigorous.
</s>