Is series $\ln\frac{n+1}{n-1}$ converging?

Question

We already know that

$$ \lim_{x\to \infty} x \ln \frac{x+1}{x-1} = 2$$

(this could be easily proved using L’Hospital’s rule.)

Then, how could we conclude (based on the limit provided) that the series

$$\sum_{n=2}^\infty \ln\frac{n+1}{n-1} $$

converges or not?

Answer 1

We know that $$\ln x>\dfrac{x-1}{x}\ \ \ \ \ \ \forall x>0,x\ne1$$ $$\implies\sum_{n=2}^\infty\ln\dfrac{n+1}{n-1}>\sum_{n=2}^\infty\dfrac{2}{n+1}=\infty$$

Edit

Proof of first inequality:

Define function $$f(x)=\ln x-\dfrac{x-1}{x}$$ $$\implies f'(x)=\dfrac{x-1}{x^2}$$ Therefore, by first derivative test, $f$ takes its minimum value at $x=1$ (verify!), which is $0$.

Answer 2

You know that $\ln \frac{n+1}{n-1}\sim \frac2n$ as $n\to\infty$, and that $\sum\frac2n$ diverges. By the limit comparison test, so does $\sum\ln \frac{n+1}{n-1}$.

Answer 3

$\sum_\limits{n=2}^N \ln \frac {n+1}{n-1}\\ \sum_\limits{n=2}^N \ln (n+1) - \ln (n-1)\\ \ln N + \ln (N+1) - \ln 2$

As $N$ approaches infinity $\ln N + \ln (N+1) - \ln 2$ also goes toward infinity.