<s>[INST] Does the following series converges or diverges? $\sum_{n=1}^{+\infty}\left[n \log\left ( \frac{2n+1}{2n-1} \right )-1\right]$ ,converges or not
My attempt
$a_n=n \log\left ( \frac{2n+1}{2n-1} \right )-1 \\ \\
\log \left ( \frac{2n+1}{2n-1} \right )=\log \left ( 1+\frac{2}{2n-1} \right ) \sim \frac{2}{2n-1} \ (n\rightarrow +\infty) \\ \\
a_n=n\log \left ( 1+\frac{2}{2n-1} \right )-1 \sim n \frac{2}{2n-1}-1=\frac{2n}{2n-1}-1=\frac{1}{2n-1}  \\ \\ $
$\sum_{n=1}^{+\infty}\frac{1}{2n-1} $  diverges  ,then $\sum_{n=1}^{+\infty}a_n $ diverges 
but according to Wolfram  $\sum_{n=1}^{+\infty}a_n=\frac{1}{2}-\frac{\log 2}{2}$
where is the mistake in my solution please?
thanks
 [/INST] $$\log\left(\frac{2n+1}{2n-1}\right) = 2\,\text{arctanh}\left(\frac{1}{2n}\right)=\frac{1}{n}+\frac{1}{12n^3}+O(n^{-5})$$
leads to
$$ n\log\left(\frac{2n+1}{2n-1}\right)-1 = \frac{1}{12n^2}+O(n^{-4}) $$
so the given series is convergent. Your initial approximation was simply not accurate enough.

The exact value of the series is simple to compute through summation by parts. We have:
$$\begin{eqnarray*} \sum_{n=1}^{N}\left[n\log\left(\frac{2n+1}{2n-1}\right)-1\right]&=&-N+\sum_{n=1}^{N}n\left[\log(2n+1)-\log(2n-1)\right]\\&=&-N+N\log(2N+1)-\sum_{n=1}^{N-1}\log(2n+1)\\&=&-N+N\log(2N+1)-\log((2N-1)!!)\end{eqnarray*}$$
so 
$$ \sum_{n=1}^{+\infty}\left[n\log\left(\frac{2n+1}{2n-1}\right)-1\right] = \frac{1-\log 2}{2}$$
follows from Stirling's inequality. Have a look at this recent question of mine.
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