I need help with a Finite Series Problem:
Find the sum to $n$ terms of
\begin{eqnarray*}
\frac{1}{1\cdot 2\cdot 3} + \frac{3}{2\cdot 3\cdot 4} + \frac{5}{3\cdot 4\cdot 5} +
        \frac{7}{4\cdot 5\cdot 6}+\cdots \\
\end{eqnarray*}
Answer:
The way I see it, the problem is asking me to find this series:
\begin{eqnarray*}
S_n &=& \sum_{i=1}^{n} {a_i} \\
\text{with } a_i &=& \frac{2i-1}{i(i+1)(i+2)} \\
\end{eqnarray*}
We have:
\begin{eqnarray*}
S_n &=& S_{n-1}  + a_n \\
S_n &=& S_{n-1}  + \frac{2n-1}{n(n+1)(n+2)} \\
\end{eqnarray*}
I am tempted to apply the technique of partial fractions
but I believe there is no closed formula for a series of the of the form:
\begin{eqnarray*}
\sum_{i=1}^{n} \frac{1}{i+k} \\
\end{eqnarray*}
where $k$ is a fixed constant. Therefore I am stuck. I am hoping that somebody
can help me.
Thanks Bob

There is more simple way (for me). You have
$$a_n=\frac{2n-1}{n(n+1)(n+2)}=-\frac52\frac{1}{n+2} + \frac{3}{n+1} - \frac{1}{2n};$$
hence
$$S_N = \sum_{n=1}^N a_n = -\frac52\left(H_{N+2}-1-\frac12\right) + 3(H_{N+1}-1) - \frac12 H_N,$$
where $H_N$ is $N$-th harmonic number. Simplify it:
$$S_N = -\frac52\left(H_N + \frac{1}{N+1} + \frac{1}{N+2}-\frac32\right) + 3\left(H_N+\frac{1}{N+1}-1\right) - \frac12H_N=\\=
\frac34 + \frac{1}{2(N+1)}-\frac{5}{2(N+2)}.$$
