Answer:
For consecutive integers, let them be \(n\) and \(n+1\). Then
\((n+1)^2-n^2=n^2+2n+1-n^2=2n+1\), which is odd.
The series is
\(\frac{3}{1^2\times 2^2}+\frac{5}{2^2\times 3^2}+\frac{7}{3^2\times 4^2}+\cdots+\frac{2r+1}{r^2(r+1)^2}+\cdots\)
For the general term, notice that
\(2r+1=(r+1)^2-r^2\).
So
\(\frac{2r+1}{r^2(r+1)^2}=\frac{(r+1)^2-r^2}{r^2(r+1)^2}=\frac{1}{r^2}-\frac{1}{(r+1)^2}\).
Hence the sum to \(n\) terms is
\(\sum_{r=1}^n \left(\frac{1}{r^2}-\frac{1}{(r+1)^2}\right)=1-\frac{1}{(n+1)^2}.\)
Therefore the sum to infinity is
\(\lim_{n\to\infty}\left(1-\frac{1}{(n+1)^2}\right)=1.\)
So the sum to \(n\) terms is \(1-\frac{1}{(n+1)^2}\), and the sum to infinity is \(1\).
Let the consecutive integers be \(n\) and \(n+1\). Their squares differ by
\((n+1)^2-n^2=n^2+2n+1-n^2=2n+1.\)
Since \(2n+1\) is of the form “even number + 1”, it is odd. So the difference between the squares of consecutive integers is an odd integer.
Now consider the series
\(\frac{3}{1^2\times 2^2}+\frac{5}{2^2\times 3^2}+\frac{7}{3^2\times 4^2}+\cdots+\frac{2r+1}{r^2(r+1)^2}+\cdots\)
The numerator in the \(r\)th term can be written as
\(2r+1=(r+1)^2-r^2.\)
So the \(r\)th term becomes
\(\frac{2r+1}{r^2(r+1)^2}=\frac{(r+1)^2-r^2}{r^2(r+1)^2}.\)
Split this into two fractions:
\(\frac{(r+1)^2}{r^2(r+1)^2}-\frac{r^2}{r^2(r+1)^2}=\frac{1}{r^2}-\frac{1}{(r+1)^2}.\)
Therefore the sum to \(n\) terms is
\(\sum_{r=1}^n \left(\frac{1}{r^2}-\frac{1}{(r+1)^2}\right).\)
Writing out the first few terms gives
\((1-\tfrac14)+(\tfrac14-\tfrac19)+(\tfrac19-\tfrac1{16})+\cdots+\left(\tfrac{1}{n^2}-\tfrac{1}{(n+1)^2}\right).\)
All the intermediate terms cancel, leaving
\(1-\frac{1}{(n+1)^2}.\)
So the sum to \(n\) terms is \(1-\frac{1}{(n+1)^2}\).
To find the sum to infinity, let \(n\to\infty\):
\(\lim_{n\to\infty}\left(1-\frac{1}{(n+1)^2}\right)=1.\)
Hence the sum to infinity is \(1\).
