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

Setting 
$$\frac{2n-1}{n(n+1)(n+2)}=\frac{An+B}{n(n+1)}-\frac{A(n+1)+B}{(n+1)(n+2)}$$
gives us $A=2,B=-\frac 12$, i.e.
$$\frac{2n-1}{n(n+1)(n+2)}=\frac{2n-\frac 12}{n(n+1)}-\frac{2(n+1)-\frac 12}{(n+1)(n+2)}.$$
Hence, we have
$$\begin{align}\sum_{i=1}^{n}\frac{2i-1}{i(i+1)(i+2)}&=\sum_{i=1}^{n}\left(\frac{2i-\frac 12}{i(i+1)}-\frac{2(i+1)-\frac 12}{(i+1)(i+2)}\right)\\&=\frac{2\cdot 1-\frac 12}{1\cdot (1+1)}-\frac{2(n+1)-\frac 12}{(n+1)(n+2)}\\&=\color{red}{\frac 34-\frac{4n+3}{2(n+1)(n+2)}}\end{align}$$
