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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!”

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    Did you try the Stirling approximation for finding an $\sim$ of $n^{s-1}-\frac{\Gamma(s+n)}{n!}$ ? – 2017-02-07
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    (+1) Nice question. Anyway, it is well-known that the values of $\zeta(s)$ in the critical strip can be computed through suitable regularizations of divergent series or integrals. – 2017-02-07
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    @JackD'Aurizio: Thanks. The term-wise regularization of zeta by gamma would shed light on the behavior of $\,\zeta’\,$ inside the critical strip from a totally different angle. Check this out: $$ \zeta'(1-s)=\sum_{n=1}^{\infty}\left[\,\frac{\Gamma(n-1+s)\,\psi(n-1+s)}{(n-1)!}-\frac{\log{n}}{n^{\small{1-s}}}\,\right] \quad\colon\,Re\{s\}\in\,(0,\,1) $$ – 2017-02-07
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    @user1952009: A term-wise asymptotic approximation with a manageable error-term would be really interesting! I barely manage to justify the whole relation with an error of order $\,\mathcal{O}\left(N^{-\sigma}\right)\,$. – 2017-02-07

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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.