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Find $\sum_{n=1}^{\infty} \frac{n^{\sigma -1} (n+\sigma )-(n+1)^{\sigma }}{\sigma(1-\sigma)}$ for $ 0<\sigma<1$

My try $ \sum_{n=1}^{\infty} \frac{n^{\sigma -1} (n+\sigma )-(n+1)^{\sigma }}{\sigma(1-\sigma)}$= $\sum_{n=1}^{\infty} \frac{n^{\sigma } -(n+1)^{\sigma }+\sigma n^{1-\sigma}}{\sigma(1-\sigma)}$ so the sum should be

$\frac{1+\sum_{n=1}^{\infty} \frac{1}{n^{1-\sigma}}}{\sigma(\sigma-1)}$. Also from here Evaluate$\int_{1}^{\infty}$ $\frac{1-(x-[x])}{x^{2-\sigma}}$dx where [x] denotes greatest integer function and $0<\sigma<1$ $\int_1^\infty(1-x+\lfloor x\rfloor )\, x^{\sigma -2}\,dx=$$ \sum_{n=1}^{\infty} \frac{n^{\sigma -1} (n+\sigma )-(n+1)^{\sigma }}{\sigma(1-\sigma)}$= $\frac{1+\sigma\sum_{n=1}^{\infty} \frac{1}{n^{1-\sigma}}}{\sigma(\sigma-1)}$. $\int_1^\infty(1-x+\lfloor x\rfloor )\, x^{\sigma -2}\,dx \leq $$\int_1^\infty$$x^{\sigma-2}$dx=$\frac{1}{1-\sigma}$. So the integral on the left in equation (1) is convergent so the summation on the right must be convergent but for$ 0<\sigma<1 , 0<1-\sigma<1$. So the series $\sum_{n=1}^{\infty} \frac{1}{n^{1-\sigma}}$ on the right is divergent. I know that $\zeta(s)$= $\sum_1^\infty \frac{1}{n^s}$ , $\Re(s)>1.$ help

Vincenzo Tibullo
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  • Your evaluation of partial sums appears to be incorrect. Prove by induction that the first $n$ terms have sum $\frac{1+\sigma\sum_{k=1}^nk^{\sigma-1}-(n+1)^\sigma}{\sigma(1-\sigma)}$. – J.G. Aug 22 '20 at 07:51
  • @J.G. Thanks. then how do we calculate the final sum?? –  Aug 22 '20 at 07:52
  • Four (!) almost-the-same questions. As said multiple times, the result uses the values of $\zeta(s)$ with $s<1$. Trying to express it in terms of "what you know" (i.e. $\zeta(s)$ for $s>1$) is more or less the same as trying to express $\zeta(3)$ in terms of $\zeta(4)$. These are simply unrelated. To get a non-trivial answer (i.e. different from "$A=A$"), and understand it, you have to learn complex analysis (particularly, the concept of analytic continuation). – metamorphy Aug 25 '20 at 14:07
  • Anyway, I've updated this answer with a "real-only" version, as complete as possible I believe. – metamorphy Aug 25 '20 at 18:11

1 Answers1

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ With $\ds{N \in \mathbb{N}_{>\ 1}}$: \begin{align} &\bbox[5px,#ffd]{\left.\sum_{n = 1}^{N} {n^{\sigma -1}\pars{n + \sigma} - \pars{n + 1}^{\sigma} \over \sigma\pars{1 - \sigma}} \,\right\vert_{\ 0\ <\ \sigma\ <\ 1}} \\[5mm] = &\ {1 \over \sigma\pars{1 - \sigma}}\bracks{% \sum_{n = 1}^{N}n^{\sigma} + \sigma\sum_{n = 1}^{N}n^{\sigma - 1} - \sum_{n = 1}^{N}\pars{n + 1}^{\sigma}} \\[5mm] = &\ {1 \over \sigma\pars{1 - \sigma}}\ \times \\[2mm] &\ \braces{% \pars{1 + \sum_{n = 2}^{N}n^{\sigma}} + \sigma\sum_{n = 1}^{N}n^{\sigma - 1} - \bracks{\sum_{n = 2}^{N}n^{\sigma} + \pars{N + 1}^{\sigma}}} \\[5mm] = & {1 \over \sigma\pars{1 - \sigma}} + {1 \over 1 - \sigma}\sum_{n = 1}^{N}{1 \over n^{1 - \sigma}} - {\pars{N + 1}^{\sigma} \over \sigma\pars{1 - \sigma}} \\[5mm] = &\ {1 \over \sigma\pars{1 - \sigma}} \\[2mm] &\ + {1 \over 1 - \sigma}\ \bracks{\zeta\pars{1 - \sigma} + {N^{\sigma} \over \sigma} + \pars{1 - \sigma}\int_{N}^{\infty}{\braces{x} \over x^{2 - \sigma}}\,\dd x} \\[2mm] &\ - {\pars{N + 1}^{\sigma} \over \sigma\pars{1 - \sigma}} \\[5mm] = & {1 + \sigma\,\zeta\pars{1 - \sigma} \over \sigma\pars{1 - \sigma}} + \int_{N}^{\infty}{\braces{x} \over x^{2 - \sigma}}\,\dd x - {\pars{N + 1}^{\sigma} - N^{\sigma} \over \sigma\pars{1 - \sigma}} \end{align} See this identity. Note that \begin{align} 0 & < \verts{\pars{1 - \sigma}\int_{N}^{\infty}{\braces{x} \over x^{2 - \sigma}}\,\dd x} < \pars{1 - \sigma}\int_{N}^{\infty}{\dd x \over x^{2 - \sigma}} \\[5mm] & = {1 \over N^{1 - \sigma}} \,\,\,\stackrel{\mrm{as}\ N\ \to \infty}{\Large\to}\,\,\, \color{red}{\large 0} \end{align} and $\ds{{\pars{N + 1}^{\sigma} - N^{\sigma} \over \sigma\pars{1 - \sigma}} \sim {1 \over 1 - \sigma} \,{1 \over N^{1 - \sigma}} \to \color{red}{0}\,\,\,}$ as $\ds{\,\,\, N \to \infty}$.

Then, $$ \bbox[5px,#ffd]{\left.\sum_{n = 1}^{\infty} {n^{\sigma -1}\pars{n + \sigma} - \pars{n + 1}^{\sigma} \over \sigma\pars{1 - \sigma}} \,\right\vert_{\ 0\ <\ \sigma\ <\ 1}} = \bbx{1 + \sigma\,\zeta\pars{1 - \sigma} \over \sigma\pars{1 - \sigma}} \\ $$

It is amusing that the solution limiting case $\ds{\sigma \to 0^{+}}$ is equal to $\ds{\gamma}$ ( the Euler-Mascheroni Constant ).
Felix Marin
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