Determine if the following series converges or diverges: $\sum _{n=1}^{\infty }\ln\left(\frac{n}{n+1}\right)$

Question

I'm having trouble understanding if the following series converges or diverges:

$$ \sum _{n=1}^{\infty }\ln\left(\frac{n}{n+1}\right) $$

I've noticed that $\lim _{x\to \infty \:}\ln \left(\frac{x}{x+1}\right) = 0$ and therefore I can't deduce that it diverges, but other than that I'm really not sure what is the right way to go here. Any help is appreciated

Answer 1

This is a telescoping sum $$ \sum _{n=1}^{\infty }\ln\left(\frac{n}{n+1}\right)=\ln(1)-\ln(2)+\ln(2)-\ln(3)+\ln(3)-\ln(4)+... $$

This means that the first condition of sum having at least one accumulation point is not fulfilled.

$$\sum _{n=1}^{m }\ln\left(\frac{n}{n+1}\right)=-\ln(m+1)$$

For this reason the series diverges.

Answer 2

HINT

Note that

$$\ln\left(\frac{n}{n+1}\right)=\ln\left(1-\frac{1}{n+1}\right)\sim-\frac 1{n+1}$$

then refer to limit comparison test.