Problem source

Show that the sum $S_N$ of the first $N$ terms of the series
$$\frac{1}{1\cdot2\cdot3}+\frac{3}{\cdot3\cdot4}+\frac{5}{3\cdot4\cdot5}+\cdots
+\frac{2n-1}{n(n+1)(n+2)}+\cdots$$
is
$${1\over2}\left({3\over2}+{1\over N+1}-{5\over N+2}\right).$$
What is the limit of $S_N$ as $N\to\infty$?
The numbers $a_n$ are such that 
$$\frac{a_n}{a_{n-1}}=\frac{(n-1)(2n-1)}{(n+2)(2n-3)}.$$
Find an expression for $a_n/a_1$  and hence, or otherwise, evaluate $\sum\limits_{n=1}^\infty a_n$ when $\displaystyle a_1=\frac{2}{9}\;$.

Solution source

First notice by partial fractions:
\begin{align*}
\frac{2n-1}{n(n+1)(n+2)} &= \frac{-1/2}{n} + \frac{3}{n+1} + \frac{-5/2}{n+2} \\
&= \frac{-1}{2n} + \frac{3}{n+1} - \frac{5}{2(n+2)}
\end{align*}

And therefore:

\begin{align*}
\sum_{n = 1}^N \frac{2n-1}{n(n+1)(n+2)}  &= -\frac12 \sum_{n=1}^N \frac1n  +3\sum_{n=1}^N \frac1{n+1} -\frac52 \sum_{n=1}^N \frac1{n+2} \\
&= -\frac12-\frac14 + \frac{3}{2}+ \sum_{n=3}^N (3-\frac12 -\frac52)\frac1n + \frac{3}{N+1} - \frac{5}{2(N+1)} - \frac{5}{2(N+2)} \\
&= \frac12 \l \frac32+\frac1{N+1}-\frac{5}{N+2} \r
\end{align*}

As $N \to \infty, S_N \to \frac{3}{4}$.

\begin{align*}
&& \frac{a_n}{a_{n-1}}&=\frac{(n-1)(2n-1)}{(n+2)(2n-3)} \\
\Rightarrow && \frac{a_n}{a_1} &= \frac{a_n}{a_{n-1}} \cdot \frac{a_{n-1}}{a_{n-2}} \cdots \frac{a_2}{a_1} \\
&&&= \frac{(n-1)(2n-1)}{(n+2)(2n-3)} \cdot \frac{(n-2)(2n-3)}{(n+1)(2n-5)} \cdots \frac{(1)(3)}{(4)(1)} \\
&&&= \frac{(2n-1)3\cdot 2\cdot 1}{(n+2)(n+1)n} \\
&&& = \frac{6(2n-1)}{n(n+1)(n+2)}
\end{align*}

Therefore $a_n = \frac{4}{3} \frac{2n-1}{n(n+1)(n+2)}$ and so our sequence is $\frac43$ the earlier sum, ie $1$