Answer: (i) The series telescopes, so the sum to \(N\) terms is

\(S_N=\frac{1}{15}-\frac{1}{(N+3)(2N+5)}\).

(ii) As \(N\to\infty\), \(\frac{1}{(N+3)(2N+5)}\to 0\), so

\(S_\infty=\frac{1}{15}\).

First rewrite the general term in a telescoping form:

\(\frac{1}{(n+2)(2n+3)}-\frac{1}{(n+3)(2n+5)}\)

Bring the two fractions together:

\(=\frac{(n+3)(2n+5)-(n+2)(2n+3)}{(n+2)(n+3)(2n+3)(2n+5)}\).

Now expand the numerator:

\((n+3)(2n+5)=2n^2+11n+15\),

\((n+2)(2n+3)=2n^2+7n+6\).

So the numerator becomes

\((2n^2+11n+15)-(2n^2+7n+6)=4n+9\).

Hence

\(\frac{1}{(n+2)(2n+3)}-\frac{1}{(n+3)(2n+5)}=\frac{4n+9}{(n+2)(n+3)(2n+3)(2n+5)}\),

as required.

Now consider

\(S_N=\sum_{n=1}^{N}\frac{4n+9}{(n+2)(n+3)(2n+3)(2n+5)}\).

Using the identity above,

\(S_N=\sum_{n=1}^{N}\left(\frac{1}{(n+2)(2n+3)}-\frac{1}{(n+3)(2n+5)}\right)\).

Write out the first few terms:

\(S_N=\left(\frac{1}{3\cdot 5}-\frac{1}{4\cdot 7}\right)+\left(\frac{1}{4\cdot 7}-\frac{1}{5\cdot 9}\right)+\cdots+\left(\frac{1}{(N+2)(2N+3)}-\frac{1}{(N+3)(2N+5)}\right).\)

Everything in the middle cancels, leaving

\(S_N=\frac{1}{3\cdot 5}-\frac{1}{(N+3)(2N+5)}=\frac{1}{15}-\frac{1}{(N+3)(2N+5)}\).

So

\(\boxed{S_N=\frac{1}{15}-\frac{1}{(N+3)(2N+5)}}\).

For the sum to infinity, let \(N\to\infty\):

\(\frac{1}{(N+3)(2N+5)}\to 0\),

therefore

\(\boxed{S_\infty=\frac{1}{15}}\).