What is the sum of this telescoping series?

Question

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

Seems like a telescoping series so everythig will cancel out except $\frac{1}{\ln 2}$ ?? is my thinking right. How do I write it formal. The series goes to infinite.

Answer 1

You can prove by induction that

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

which means that

$$\sum_{n=1}^\infty \frac{1}{\ln(n+2)} - \frac{1}{\ln(n+1)} \\= \lim_{N\to\infty}\left(\sum_{n=1}^N \frac{1}{\ln(n+2)} - \frac{1}{\ln(n+1)}\right) \\= \lim_{N\to\infty}\frac{1}{\ln(N+2)} -\frac{1}{\ln 2}$$

which should be easy to calculate.

Answer 2

You can write it to an index $N$ to obtain $\frac{1}{\ln(N+2)} - \frac{1}{\ln(2)}$ and then tend $N$ to infinity