Prove $\frac {1}{1\cdot2} - \frac {1}{2\cdot3} + \frac{1}{3\cdot4}-\cdots = \ln 2 - 1$

Question

Please help me with this problem.

I have figured out the $n^{th}$ term as $\frac {(-1)^{n+1}}{n(n+1)}$. How do I proceed ?

Answer 1

Hint: Use the fact that $$\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$$ and that $$\ln2=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}.$$ Also, as stated, the right hand side should be $2\ln 2-1$, not $\ln2 -1$