Proving $\sum_{i=1}^{n} {\frac{1}{4i^2-1}} = \frac{n}{2n +1}$ for all $n \in N$ using mathematical induction I'm having a horrible time completing this since this is the first time I'm learning mathematical induction.
So far I have this:
Let $P(n)$ be:
$P(n):\sum_{i=1}^{n} {\frac{1}{4i^2-1}} = \frac{n}{2n +1}$
For the base case, $n=1$
$LHS = \sum_{i=1}^{0} {\frac{1}{4i^2-1}} = {\frac{1}{4(1^2)-1}} = {\frac{1}{4-1}} = {\frac{1}{3}}$
$RHS = \frac{n}{2n +1} = \frac{1}{2(1) +1} = \frac{1}{2 +1} = \frac{1}{3}$
Therefore $P(1)$ holds.
Assuming $P(k)$, $P(k)$ holds for $n=k$:
$P(k): \sum_{i=1}^{k} {\frac{1}{4i^2-1}} = \frac{k}{2k +1}$
Proving for $P(k+1)$:
$P(k+1):\\ \\\ \\ \sum_{i=1}^{k+1} {\frac{1}{4i^2-1}}$
$= \sum_{i=1}^{k} {\frac{1}{4i^2-1}} + \frac{1}{4(k+1)^2-1}$
$=\frac{k}{2k +1} + \frac{1}{4(k+1)^2-1}$
Can anyone help me from this point onward?

Note that the difference between the $(n+1)$th and $n$th putative partial sums is $$\begin{align*}\frac{n+1}{2(n+1)+ 1} - \frac{n}{2n+1} &= \frac{(n+1)(2n+1) - n (2n+3)}{(2n+3)(2n+1)} \\
&= \frac{1}{4n^2 + 8n + 3} \\
&= \frac{1}{4 (n+1)^2 - 1}\end{align*}$$
and this is the $(n+1)$th term in the series.
