Find $\sum_{n=1}^{\infty} \frac{n^{\sigma -1} (n+\sigma )-(n+1)^{\sigma }}{\sigma(1-\sigma)}$ for $ 0<\sigma<1$ 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
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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 ).
