I am investigating the convergence of $$\sum _{n=1}^{\infty }\left\{ 1-n\log \frac {2n+1} {2n-1}\right\} $$
Now as per Cauchy's test for absolute convergence.
If $\overline {\lim _\limits{n\rightarrow \infty }}\left| u_{n}\right|^{{1}/{n}} < 1,\sum _\limits{n=1}^{\infty }u_{n}$ converges absolutely
Obviously, if $\overline {\lim \limits_{n\rightarrow \infty }}\left| u_{n}\right|^{{1}/{n}} > 1,\sum _\limits{n=1}^{\infty }u_{n}$ does not converge.
I observed $$\overline {\lim _{n\rightarrow \infty }}\left| \log \left( \dfrac {2n+1} {2n-1}\right) ^{-n}\right| = \overline {\lim _{n\rightarrow \infty }}\left| \log \left( 1-\dfrac {1} {n-{1}/{2}}\right) ^{-n}\right|$$
Could I take the negative power of $n$ outside the absolute brackets here? I guess even if I could establish $\log$ part converges that would only show that the overall series diverges right. Is that the correct result ? Any alternative lines of attacking this problem would be much appreciated.
Best Answer
You could prove the convergence of the series using a comparison criterion. For example, calculate the limit of $|a_n|/(\frac{1}{n^2})$. You should then calculate $$\lim_{n \to \infty}\left| n^2-n^3\log \left(\frac{2n+1}{2n-1} \right)\right| $$.
For this calculation, the simplest method I could think of was expanding in Taylor series.
$$\log(x+1)-\log(x-1)=\log \left(1+\frac{1}{x}\right)-\log\left(1-\frac{1}{x}\right)=2\sum_{k \text{ odd}}\frac{1}{kx^k}$$
Then you have to calculate
$$ \lim_{n \to \infty} \left|n^2-n^3\cdot 2\left(\frac{1}{2n}+\frac{1}{3(2n)^3}+\frac{1}{5(2n)^5}+... \right)\right|=\frac{1}{12}$$
Therefore $\sum a_n$ is absolutely convergent, and in particular convergent.