Answer: We first write
\(\dfrac{1}{(2r+1)(2r+3)}=\dfrac12\left(\dfrac{1}{2r+1}-\dfrac{1}{2r+3}\right).\)
Hence
\(\sum_{r=1}^{N}\dfrac{1}{(2r+1)(2r+3)}=\dfrac12\sum_{r=1}^{N}\left(\dfrac{1}{2r+1}-\dfrac{1}{2r+3}\right).\)
Writing out the terms,
\(\dfrac12\left(\dfrac13-\dfrac15\right)+\dfrac12\left(\dfrac15-\dfrac17\right)+\cdots+\dfrac12\left(\dfrac{1}{2N+1}-\dfrac{1}{2N+3}\right).\)
The middle terms cancel, so
\(\sum_{r=1}^{N}\dfrac{1}{(2r+1)(2r+3)}=\dfrac12\left(\dfrac13-\dfrac{1}{2N+3}\right)=\dfrac16-\dfrac{1}{2(2N+3)}.\)
Now
\(\sum_{r=N+1}^{2N}\dfrac{1}{(2r+1)(2r+3)}=\sum_{r=1}^{2N}\dfrac{1}{(2r+1)(2r+3)}-\sum_{r=1}^{N}\dfrac{1}{(2r+1)(2r+3)}.\)
Using the result above,
\(\sum_{r=N+1}^{2N}\dfrac{1}{(2r+1)(2r+3)}=\left(\dfrac16-\dfrac{1}{2(4N+3)}\right)-\left(\dfrac16-\dfrac{1}{2(2N+3)}\right).\)
So
\(\sum_{r=N+1}^{2N}\dfrac{1}{(2r+1)(2r+3)}=\dfrac12\left(\dfrac{1}{2N+3}-\dfrac{1}{4N+3}\right)=\dfrac{N}{(2N+3)(4N+3)}.\)
Since \((2N+3)(4N+3)\gt (2N)(4N)\),
\(\dfrac{N}{(2N+3)(4N+3)}\lt \dfrac{N}{(2N)(4N)}=\dfrac{1}{8N}.\)
Therefore
\(\sum_{r=N+1}^{2N}\dfrac{1}{(2r+1)(2r+3)}\lt \dfrac{1}{8N}.\)
Let
\(S_N=\sum_{r=1}^{N}\dfrac{1}{(2r+1)(2r+3)}.\)
Use partial fractions:
\(\dfrac{1}{(2r+1)(2r+3)}=\dfrac{A}{2r+1}+\dfrac{B}{2r+3}.\)
Then
\(1=A(2r+3)+B(2r+1).\)
Comparing coefficients gives \(2A+2B=0\), so \(B=-A\). Also \(3A+B=1\), hence \(3A-A=1\), so \(2A=1\) and \(A=\dfrac12\). Therefore \(B=-\dfrac12\).
So
\(\dfrac{1}{(2r+1)(2r+3)}=\dfrac12\left(\dfrac{1}{2r+1}-\dfrac{1}{2r+3}\right).\)
Hence
\(S_N=\dfrac12\sum_{r=1}^{N}\left(\dfrac{1}{2r+1}-\dfrac{1}{2r+3}\right).\)
Writing out the sum:
\(S_N=\dfrac12\left(\left(\dfrac13-\dfrac15\right)+\left(\dfrac15-\dfrac17\right)+\cdots+\left(\dfrac{1}{2N+1}-\dfrac{1}{2N+3}\right)\right).\)
This is telescoping, so all interior terms cancel and
\(S_N=\dfrac12\left(\dfrac13-\dfrac{1}{2N+3}\right)=\dfrac16-\dfrac{1}{2(2N+3)}.\)
This shows that
\(\sum_{r=1}^{N}\dfrac{1}{(2r+1)(2r+3)}=\dfrac16-\dfrac{1}{2(2N+3)}.\)
Now let
\(T=\sum_{r=N+1}^{2N}\dfrac{1}{(2r+1)(2r+3)}.\)
Then
\(T=\sum_{r=1}^{2N}\dfrac{1}{(2r+1)(2r+3)}-\sum_{r=1}^{N}\dfrac{1}{(2r+1)(2r+3)}.\)
Applying the formula for the partial sum gives
\(T=\left(\dfrac16-\dfrac{1}{2(4N+3)}\right)-\left(\dfrac16-\dfrac{1}{2(2N+3)}\right).\)
Therefore
\(T=\dfrac12\left(\dfrac{1}{2N+3}-\dfrac{1}{4N+3}\right).\)
Combining the fractions,
\(T=\dfrac12\cdot\dfrac{(4N+3)-(2N+3)}{(2N+3)(4N+3)}=\dfrac12\cdot\dfrac{2N}{(2N+3)(4N+3)}=\dfrac{N}{(2N+3)(4N+3)}.\)
Now \(2N+3\gt 2N\) and \(4N+3\gt 4N\), so
\((2N+3)(4N+3)\gt (2N)(4N).\)
Since all quantities are positive,
\(T=\dfrac{N}{(2N+3)(4N+3)}\lt \dfrac{N}{(2N)(4N)}=\dfrac{1}{8N}.\)
Hence
\(\sum_{r=N+1}^{2N}\dfrac{1}{(2r+1)(2r+3)}\lt \dfrac{1}{8N}.\)
