Answer: (i) \(\displaystyle \sum_{r=1}^{N}\frac{1}{(3r+1)(3r-2)}=\frac13-\frac{1}{3(3N+1)}\).
(ii) \(\displaystyle \lim_{N\to\infty}\sum_{r=N+1}^{N^2}\frac{N}{(3r+1)(3r-2)}=\frac19\).
(i) First use partial fractions:
\(\dfrac{1}{(3r+1)(3r-2)}=\dfrac13\left(\dfrac{1}{3r-2}-\dfrac{1}{3r+1}\right)\).
Therefore
\(\displaystyle \sum_{r=1}^{N}\dfrac{1}{(3r+1)(3r-2)}=\dfrac13\sum_{r=1}^{N}\left(\dfrac{1}{3r-2}-\dfrac{1}{3r+1}\right)\).
Writing out the terms,
\(\displaystyle \dfrac13\left(1-\dfrac14+\dfrac14-\dfrac17+\cdots+\dfrac{1}{3N-2}-\dfrac{1}{3N+1}\right)\).
All the middle terms cancel, leaving
\(\displaystyle \dfrac13\left(1-\dfrac{1}{3N+1}\right)=\dfrac13-\dfrac{1}{3(3N+1)}\),
as required.
(ii) Use the result from part (i) twice:
\(\displaystyle \sum_{r=N+1}^{N^2}\frac{N}{(3r+1)(3r-2)}=N\left(\sum_{r=1}^{N^2}\frac{1}{(3r+1)(3r-2)}-\sum_{r=1}^{N}\frac{1}{(3r+1)(3r-2)}\right)\).
So
\(\displaystyle =N\left(\frac13-\frac{1}{3(3N^2+1)}-\frac13+\frac{1}{3(3N+1)}\right)\)
\(\displaystyle =\frac{N}{3(3N+1)}-\frac{N}{3(3N^2+1)}\).
Putting over a common denominator gives
\(\displaystyle \frac{N^3-N^2}{(3N+1)(3N^2+1)}\).
Hence, as \(N\to\infty\), the limit is
\(\displaystyle \frac{1}{9}\).
