Expressing Zeta function using Gamma series Motivated by Gautschi double inequality, 
$$ \frac{n^{s}}{n^{\small1}}\ge\frac{\Gamma(n+s)}{\Gamma(n+1)}\ge\frac{(n+1)^{s}}{(n+1)^{\small1}}\ge\frac{\Gamma(n+1+s)}{\Gamma(n+1+1)}\ge\,\cdots \quad\colon\,0\lt{s}\lt1\tag{1} $$ 
From the main definition of zeta function, 
$$ \zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}\,\,\,\colon\,Re\{s\}\gt1 \space\Rightarrow\space \zeta(1-s)=\sum_{n=1}^{\infty}\frac{n^{s}}{n^{\small1}} \qquad\colon\,Re\{s\}\lt0\tag{2} $$ 
And the sum identity of gamma function, 
$$ \sum_{n=0}^{\infty}\frac{\Gamma(n+s)}{n!}=0 \quad\Rightarrow\quad \Gamma(s)=-\sum_{n=1}^{\infty}\frac{\Gamma(n+s)}{\Gamma(n+1)} \qquad\colon\,Re\{s\}\lt0\tag{3} $$ 

How to Prove, Disprove, or Justify: 
  $$ \zeta(1-s)+\Gamma(s)=\sum_{n=1}^{\infty}\left[\,\frac{n^{s}}{n^{\small1}}-\frac{\Gamma(n+s)}{\Gamma(n+1)}\,\right] \qquad\qquad\colon\,0\lt{s}\lt1\tag{4} $$ 
  Does it hold if extended to complex plan $\,s\in\mathbb C\,$ inside the critical strip $\small 0\lt Re\{s\}\lt1\,$ ? 

“By the monotonic decreasing behavior of the inequality, the subtracting result of the two divergent series converge one step forward, covering the critical strip!”
 A: My first thought is to give an integral representation for the general term of the series, $$\frac{1}{n^{1-s}}-\frac{B(n+s,1-s)}{\Gamma(1-s)}=\frac{1}{\Gamma(1-s)}\left(\int_{0}^{+\infty}x^{-s}e^{-n x}\,dx-\int_{0}^{1}x^{-s}(1-x)^{n+s-1}\,dx\right)$$
$$\frac{1}{n^{1-s}}-\frac{B(n+s,1-s)}{\Gamma(1-s)}=\frac{1}{\Gamma(1-s)}\left(\int_{0}^{+\infty}x^{-s}e^{-n x}\,dx-\int_{0}^{+\infty}(1-e^{-x})^{-s}e^{-(n+s)x}\,dx\right)$$
Summing over $n\geq 1$ we get:
$$\sum_{n\geq 1}\left(\frac{n^s}{n^1}-\frac{\Gamma(n+s)}{\Gamma(n+1)}\right)=\frac{1}{\Gamma(1-s)}\int_{0}^{+\infty}\left(\frac{1}{x^s}-\frac{1}{(e^x-1)^s}\right)\frac{dx}{e^x-1}$$
for every $s$ with real part $\in(0,1)$. The explicit computation of the last integral as $\,\frac{\pi}{\sin(\pi s)}+\Gamma(1-s)\,\zeta(1-s)\,$ proves OP's identity $(4)$. For the computation we may use, for instance, the classical application of Ramanujan's master theorem to Bernoulli polynomials.
A: That the identity (4) is true for $\Re(s) < 0$ and that its RHS converges and stays meromorphic for $\Re(s)<1$ implies it is true for $\Re(s) < 1$.
That it converges and is meromorphic is a consequence of Jack D'Aurizio's beta function representation $$\frac{n^s}{n^1}-\frac{\Gamma(n+s)}{\Gamma(n+1)}=\frac{1}{\Gamma(1-s)}\int_{0}^\infty x^{-s}(1- (\frac{e^x-1}{x})^{-s})e^{-n x}dx\tag{5}$$
