Problem 42 (difficulty: 3/10)

Prove that the following identity holds for all positive integer \(\displaystyle n\):

\(\displaystyle \frac{1}{1\cdot 3} +\frac{1}{3\cdot 5}+\ldots + \frac{1}{(2n-1)\cdot (2n+1)}=\frac{n}{2n+1}.\)

Solution:

Induction on \(\displaystyle n\). For \(\displaystyle n=1\) we have \(\displaystyle \frac1{1\cdot3}=\frac13\) \(\displaystyle \surd\). Suppose now that the identity holds for \(\displaystyle n\), then for \(\displaystyle n+1\) we have L.H.S.

Give me another random problem!