Answer: \(\sum_{r=1}^{n}\frac1{(2r)^2-1}=\frac{n}{2n+1}\) for all positive integers \(n\). Also, \(\sum_{r=1}^{\infty}\frac1{(2r)^2-1}=\frac12\).

Let \(H_n\) be the statement

\(\displaystyle \sum_{r=1}^{n} \frac{1}{(2r)^2-1}=\frac{n}{2n+1}.\)

We prove \(H_n\) by induction.

Base case: \(n=1\)

\(\displaystyle \sum_{r=1}^{1} \frac{1}{(2r)^2-1}=\frac{1}{2^2-1}=\frac{1}{3}.\)

Also \(\displaystyle \frac{1}{2\cdot 1+1}=\frac{1}{3}.\) So \(H_1\) is true.

Inductive step

Assume that for some positive integer \(k\),

\(\displaystyle \sum_{r=1}^{k} \frac{1}{(2r)^2-1}=\frac{k}{2k+1}.\)

We must show that

\(\displaystyle \sum_{r=1}^{k+1} \frac{1}{(2r)^2-1}=\frac{k+1}{2(k+1)+1}.\)

Starting from the inductive hypothesis,

\(\displaystyle \sum_{r=1}^{k+1} \frac{1}{(2r)^2-1}=\sum_{r=1}^{k} \frac{1}{(2r)^2-1}+\frac{1}{(2k+2)^2-1}.\)

So

\(\displaystyle \sum_{r=1}^{k+1} \frac{1}{(2r)^2-1}=\frac{k}{2k+1}+\frac{1}{(2k+2)^2-1}.\)

Now factorise the new denominator:

\(\displaystyle (2k+2)^2-1=4k^2+8k+3=(2k+1)(2k+3).\)

Hence

\(\displaystyle \frac{k}{2k+1}+\frac{1}{(2k+1)(2k+3)}\)

\(\displaystyle =\frac{k(2k+3)+1}{(2k+1)(2k+3)}\)

\(\displaystyle =\frac{2k^2+3k+1}{(2k+1)(2k+3)}.\)

Factorising the numerator gives

\(\displaystyle 2k^2+3k+1=(2k+1)(k+1),\)

so

\(\displaystyle \sum_{r=1}^{k+1} \frac{1}{(2r)^2-1}=\frac{(2k+1)(k+1)}{(2k+1)(2k+3)}=\frac{k+1}{2k+3}.\)

This is exactly

\(\displaystyle \frac{k+1}{2(k+1)+1}.\)

Therefore \(H_k \Rightarrow H_{k+1}\).

Since \(H_1\) is true and \(H_k \Rightarrow H_{k+1}\), it follows by mathematical induction that

\(\displaystyle \sum_{r=1}^{n} \frac{1}{(2r)^2-1}=\frac{n}{2n+1}\)

for all positive integers \(n\).

For the infinite sum, use the formula for the partial sums:

\(\displaystyle \sum_{r=1}^{\infty} \frac{1}{(2r)^2-1}=\lim_{n\to\infty} \frac{n}{2n+1}=\frac{1}{2}.\)