Prove that $\displaystyle\sum\limits_{i=1}^{k}\left(\dfrac{1}{(2i-1)}\dfrac{1}{(2i+1)}\right) = \dfrac{k}{(2k+1)}$
My Base of Induction is to check that it is true for i=1, so:
$\dfrac{1}{1}\dfrac{1}{3} = \dfrac{1}{3}$. TRUE.
I check also if it is true for i=2.
$\dfrac{1}{3} + \dfrac{1}{3}\dfrac{1}{5} = \dfrac{6}{15}= \dfrac{2}{5}$. So it is also true.
My induction hypothesis is assume that it is true for i=k and then add up the (k+1) term. \begin{align*} \dfrac{k}{2k+1} + \left(\dfrac{1}{2k+2-1}\dfrac{1}{2k+2+1}\right)=\\ \dfrac{k}{2k+1} + \left(\dfrac{1}{2k+1}\dfrac{1}{2k+3}\right)=\\ \dfrac{k}{2k+1} + \left[\dfrac{2k+3}{(2k+1)(2k+3)}\dfrac{2k+1}{(2k+1)(2k+3)}\right]=\\ \dfrac{k}{2k+1} + \left[\dfrac{4k+4}{(2k+1)(2k+3)}\right]=\\ \dfrac{k(2k+3)}{(2k+1)(2k+3)} + \left[\dfrac{4k+4}{(2k+1)(2k+3)}\right]=\\ \dfrac{(2k^2+3k)}{(2k+1)(2k+3)} + \left[\dfrac{4k+4}{(2k+1)(2k+3)}\right] \end{align*}
But I don´t know how to go any further. What am I missing here?
$$\begin{align*} \dfrac{k}{2k+1} + \left(\dfrac{1}{2k+2-1}\dfrac{1}{2k+2+1}\right)=\\ \dfrac{k}{2k+1} + \left(\dfrac{1}{2k+1}\dfrac{1}{2k+3}\right)=\\ \frac{k(2k+3)}{(2k+1)(2k+3)} + \frac{1}{(2k+1)(2k+3)}=\\ \dfrac{k(2k+3)+1}{(2k+1)(2k+3)}=\\ \dfrac{2k^2+3k+1}{(2k+1)(2k+3)}=\\ \dfrac{(2k+1)(k+1)}{(2k+1)(2k+3)}=\\ \end{align*}$$
Hopefully you can complete from here