I have a series like this:
$$ \sum\limits_{m=0}^{n-1}\frac{2}{n(2n - 2m -1)} $$
In the end, it should be a function of $n$ only where $m$ should be represented by $n$ somehow, I really cannot remember which series should I use to calculate the result. I imagine this series should converge at least, right? if so, what is the final result.
Thanks!
$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\color{#66f}{\large% \sum_{m\ =\ 0}^{n - 1}{2 \over n\pars{2n - 2m - 1}}} =-\,{1 \over n}\sum_{m\ =\ 0}^{n - 1}{1 \over m - n + 1/2} \\[5mm]&=-\,{1 \over n}\sum_{m\ =\ 0}^{\infty} \pars{{1 \over m - n + 1/2} - {1 \over m + 1/2}} =-\sum_{m\ =\ 0}^{\infty}{1 \over \pars{m + 1/2}\pars{m - n + 1/2}} \\[5mm]&=-\,{\Psi\pars{1/2} - \Psi\pars{1/2 - n}\over 1/2 - \pars{1/2 - n}} ={\Psi\pars{1/2 - n} - \Psi\pars{1/2} \over n} \end{align} where $\ds{\Psi}$ is the Digamma function.
Moreover, $\ds{\Psi\pars{x} = H_{x - 1} - \gamma}$. $\ds{H_{x}}$ is the Harmonic Number and $\ds{\gamma}$ is the Euler-Mascheroni constant. Then,
$$ \color{#66f}{\large% \sum_{m\ =\ 0}^{n - 1}{2 \over n\pars{2n - 2m - 1}}} =\color{#66f}{\large{H_{-1/2 - n} - H_{-1/2} \over n}} $$
Another useful relation is found with:
\begin{align} \Psi\pars{\half - n} - \Psi\pars{\half}&= \braces{\Psi\pars{\half + n} + \pi\cot\pars{\pi\bracks{n + \half}}} -\bracks{-\gamma - 2\ln\pars{2}} \\[5mm]&=\Psi\pars{{3 \over 2} + n} - {1 \over n + 1/2} + \gamma + 2\ln\pars{2} \\[5mm]&=H_{n + 1/2} - {2 \over 2n + 1} + 2\ln\pars{2} ={\pars{2n + 1}\bracks{H_{n + 1/2} + 2\ln\pars{2}} - 2 \over 2n + 1} \end{align}
such that
$$ \color{#66f}{\large% \sum_{m\ =\ 0}^{n - 1}{2 \over n\pars{2n - 2m - 1}}} =\color{#66f}{\large{\pars{2n + 1}\bracks{H_{n + 1/2} + 2\ln\pars{2}} - 2 \over n\pars{2n + 1}}} $$
