Investigate the convergence of the series $\sum_{n\geq 1} a_n, \; a_n =n\ln \frac{(2n+1)}{(2n-1)}-1$ Investigate the convergence of the series $$\sum_{n\geq 1} a_n,\qquad  a_n =n\ln \frac{(2n+1)}{(2n-1)}-1$$
I tried using d'Alembert's ratio test to solve this problem, but I got this nasty polynomial I don't know what to do with: $$\frac{ a_{n+1}}{a_n} = \frac{\left(1+\frac 1n\right)\ln\left(1+\frac{2}{2n+1}\right)-\frac 1n}{\ln\left(1+\frac{2}{2n-1}\right)-\frac 1n}$$
I would be very grateful for any tips or hints regarding my problem. 
 A: hint
$$\frac {2n+1}{2n-1}=1+\frac {2}{2n-1} $$
$$\ln (1+X)=X-\frac {X^2}{2}(1+\epsilon (X)) $$
$$a_n=\frac {2n}{2n-1}-\frac {2n}{(2n-1)^2}(1+\epsilon (n))-1$$
$$\sim \frac {-1}{(2n-1)^2} \;\;(n\to +\infty) $$
thus, $\sum a_n $ converges.
A: The author says in the comment "I do know that the series diverges (tends to 0)". 
His statement shows he knows nothing about the series!
And in fact the series converges. When n tends to infinity, for this expression you have to evaluate limits involving indeterminate forms (since you have $\frac{0}{0}$ indeterminance) - use L'Hopital's rule.
A: We can do much more than showing the given series is convergent. We have
$$ \sum_{n\geq 1}\left(n \log\frac{2n+1}{2n-1}-1 \right)=\sum_{n\geq 1}\left(2n\,\text{arctanh}\frac{1}{2n}-1\right)=\sum_{n\geq 1}\sum_{m\geq 1}\frac{2}{(2m+1)(2n)^{2m}} $$
hence the given series equals:
$$ \sum_{m\geq 1}\frac{\zeta(2m)}{2^{2m}(2m+1)}=\int_{0}^{\frac{1}{2}}\left(1-\pi x \cot(\pi x)\right)\,dx \stackrel{\text{IBP}}{=} \color{red}{\frac{1-\log 2}{2}}.$$
Convergence simply follows from squeezing and the $p$-test:
$$\forall n\geq 1,\qquad  0\leq \left(2n\,\text{arctanh}\frac{1}{2n}-1\right) \leq \frac{1}{10n^2}$$
leading to $1-\log(2)\leq\frac{\pi^2}{30}\leq\frac{1}{3}$, for instance.
A: @Jack D'Aurizio beat me to it, but I just derived the same result in another way that may be interesting, too: The $n$th partial sum is
$$S_n=\sum^n_{k=1}n\,[\ln(2\,k+1)-\ln(2\,k-1)]-n.$$ Using partial summation, we obtain $$S_n=(n+1)\,\ln(2\,n+1)-\sum^n_{k=1}\ln(2\,k+1)-n,$$ this can be rewritten as $$S_n=(n+1)\,\ln(2\,n+1)-\ln(2\,n+1)!-\ln n!-n\,\ln 2-n.$$ With Stirlings approximation for the factorials, this becomes
$$S_n=\frac{1}{2}\ln\frac{n}{2\,n+1}+n\,\ln\frac{2\,n}{2\,n+1}+1+o(1),$$ so the limit as $n\rightarrow\infty$ exists and is $$-\frac{1}{2}\ln 2-\frac{1}{2}+1=\frac{1-\ln 2}{2}.$$ Before that, I had computed the sum numerically (the series doesn't converge very quickly, but there are tricks to accelerate that), and I was content that both results agreed. Well, the first 11 decimals agreed, that is satisfactory. :)
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\sum_{n = 1}^{\infty}\bracks{n\ln\pars{2n + 1 \over 2n - 1} - 1}  =
\sum_{n = 1}^{\infty}\braces{n\bracks{%
\ln\pars{1 + {1 \over 2n}} - \ln\pars{1 - {1 \over 2n}}} - 1}
\\[5mm] = &\
\sum_{n = 1}^{\infty}\int_{0}^{1}\pars{{n \over x + 2n} - {n \over x - 2n} - 1}
\dd x =
{1 \over 4}\int_{0}^{1}x\sum_{n = 0}^{\infty}
\pars{{1 \over n + 1- x/2} - {1 \over n + 1 + x/2}}\dd x
\\[5mm] = &\
{1 \over 4}\int_{0}^{1}x\bracks{%
2\,\partiald{\ln\pars{\Gamma\pars{1 + x/2}}}{x} +
2\,\partiald{\ln\pars{\Gamma\pars{1 - x/2}}}{x}}\,\dd x
\\[5mm] = &\
{1 \over 2}\int_{0}^{1}x
\,\partiald{\bracks{\ln\pars{\Gamma\pars{1 + x/2}}\Gamma\pars{1 - x/2}}}{x}
\,\dd x
\\[5mm] & =
{1 \over 2}\,\ln\pars{\Gamma\pars{3 \over 2}\Gamma\pars{1 \over 2}} -
{1 \over 2}\int_{0}^{1}\ln\pars{{x \over 2}\,{\pi \over \sin\pars{\pi x/2}}}
\\[5mm] = &\
{1 \over 2}\,\ln\pars{\pi \over 2} -
{1 \over 2}\bracks{\ln\pars{\pi \over 2} + \int_{0}^{1}\ln\pars{x}\,\dd x -
{2 \over \pi}\int_{0}^{\pi/2}\ln\pars{\sin\pars{x}}\,\dd x}
\\[5mm] = &\
-\,{1 \over 2}\bracks{-1 + \ln\pars{2}} = \bbx{1 - \ln\pars{2} \over 2}\quad
\mbox{because}\quad
\pars{\substack{\ds{\int_{0}^{1}\ln\pars{x}\,\dd x = - 1}\\[3mm]
                \ds{\underbrace{\int_{0}^{\pi/2}\ln\pars{\sin\pars{x}}\,\dd x =
                    -\,{\pi \over 2}\,\ln\pars{2}}_{frequent\ question\ !!!}}
}}
\end{align}
