Proving sum of series $\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\dots+\frac{1}{(2n-1)(2n+1)}=\frac{n}{2n+1}$

Question

I am trying to prove by induction that the sum $S_n$ of the first $n$ terms of the series $$\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\dots +\frac{1}{(2r-1)(2r+1)}+\dots $$ is $\frac{n}{2n+1}$. I have said let $n = k$. Then

\begin{align*} S_{k+1}& =S_k+ \frac{1}{(2[k+1]-1)(2[k+1]+1)} \\ &=\frac{k}{2k+1}+ \frac{1}{(2k+1)(2k+3)} \\ &=\frac{k(2k+1)(2k+3)+(2k+1)}{(2k+1)(2k+1)(2k+3)}\\ & =\frac{k(2k+3)+1}{(2k+1)(2k+3)} \\ &=\frac{2k^2+3k+1}{4k^2+8k+3}, \end{align*} but this is not of the form $\frac{n}{2n+1}$.

Answer 1

We have$$\frac{2k^2 + 3k + 1}{4k^2 + 8k + 3} = \frac{(2k+1)(k+1)}{(2k+1)(2k+3)}.$$ The rest should be straightforward.