Nature of $\zeta(s)-\zeta(s-1)$

Question

I am trying to study the nature of the $\zeta(s)-\zeta(s-1)$ function for real $s$ and larger $s$ too. I am unable to find a closed form despite my attempts of using the sum form of the $\zeta$ function and finding no good results that help me to study the nature of the function. The only result I was able to get, despite my attempts, was just $$\zeta(s)-\zeta(s-1)=\sum_{n=1}^{\infty}\frac{n^{s-1}(n-1)}{n^{2s-1}}$$ which did not help me to study the nature of the function at all. Can anybody help me proceed?

Answer 1

Introduction

For the following proof, you will require prerequisite knowledge about the $\Gamma$ function and its integral form. I will aim to find a representation for the function you are interested in, $\zeta(s)-\zeta(s-1)$, and find its nature for larger real $s$.

Procedure

A representation for the $\zeta$ function

\begin{align} \zeta(s)=\sum_{n\geq 1}\frac{1}{n^s}=\frac{1}{\Gamma(s)}\sum_{n\geq 1}\frac{1}{n^s}\int_0^{\infty}e^{-x}x^{s-1}\mathrm{d}x \end{align} If we substitute $x=nv$ in the above equation for some $v$ then we arrive at the following equation; \begin{align} \zeta(s)&=\frac{1}{\Gamma(s)}\sum_{n\geq 1}\frac{1}{n^s}\int_0^{\infty}e^{-nv}n^{s-1}v^{s-1}\cdot n\mathrm{d}v\\ &=\frac{1}{\Gamma(s)}\sum_{n\geq 1}\int_0^{\infty}e^{-nv}v^{s-1}\mathrm{d}v\\ &=\frac{1}{\Gamma(s)}\int_0^{\infty}\sum_{n\geq 1}e^{-nv}v^{s-1}\mathrm{d}v\\ &=\frac{1}{\Gamma(s)}\int_0^{\infty}\frac{v^{s-1}}{e^v-1}\mathrm{d}v\\ &=\frac{1}{\Gamma(s)}\int_0^{\infty}\frac{x^{s-1}}{e^x-1}\mathrm{d}x\\ &=\frac{1}{s\Gamma(s)}\int_0^{\infty}\frac{x^se^x}{(e^x-1)^2}\mathrm{d}x \end{align} From the last equation, if we change $s\to s-1$ then we get the following $\because s\Gamma(s)=\Gamma(s+1)$: \begin{align} \zeta(s-1)=\frac{1}{\Gamma(s)}\int_0^{\infty}\frac{x^{s-1}e^x}{(e^x-1)^2}\mathrm{d}x \end{align}

The function $\zeta(s)-\zeta(s-1)$

\begin{align} \zeta(s)-\zeta(s-1)&=\frac{1}{\Gamma(s)}\int_0^{\infty}\frac{x^{s-1}}{e^x-1}\mathrm{d}x-\frac{1}{\Gamma(s)}\int_0^{\infty}\frac{x^{s-1}e^x}{(e^x-1)^2}\mathrm{d}x\\ &=\frac{1}{\Gamma(s)}\left[\int_0^{\infty}\frac{x^{s-1}}{e^x-1}\mathrm{d}x-\int_0^{\infty}\frac{x^{s-1}e^x}{(e^x-1)^2}\mathrm{d}x\right]\\ &=\frac{1}{\Gamma(s)}\int_0^{\infty}\frac{x^{s-1}(e^x-1)-x^{s-1}e^x}{(e^x-1)^2}\mathrm{d}x\\ &=-\frac{1}{\Gamma(s)}\int_0^{\infty}\frac{x^{s-1}}{(e^x-1)^2}\mathrm{d}x \end{align}

Behavior for larger $s$

As $s\to\infty$ it is easy to notice that $\frac{x^{s-1}}{\Gamma(s)}\to 0$, even by looking at the infinite series for $e$, it helps give an intuition for this fact, however it concludes that this function $\zeta(s)-\zeta(s-1)\to 0$ considering only real values of $s$.