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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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Hazem Orabi
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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. – Jack D'Aurizio Feb 07 '17 at 18:56
  • @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 $,\mathcal{O}\left(N^{-\sigma}\right),$. – Hazem Orabi Feb 07 '17 at 20:24
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    The question was answered 3 years ago, what do you want more. – reuns Jan 23 '20 at 14:18
  • @reuns: I just wanted to draw more attention to this direct connection between the power behavior represented by zeta function and the factorial behavior represented by gamma function inside the critical strip. Many thanks anyway. – Hazem Orabi Jan 26 '20 at 13:29

2 Answers2

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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.

Jack D'Aurizio
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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 $$\frac{n^s}{n^1}-\frac{\Gamma(n+s)}{\Gamma(n+1)}=\frac{1}{\Gamma(1-s)}\int_{0}^\infty x^{-s}(1- (\frac{e^x-1}{x})^{-s})e^{-n x}dx\tag{5}$$

reuns
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