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

You can write it as $\sum_{k\ge0}\frac{2k+1}{(k+3)_3}=\sum_{k\ge0}(2k+1)(k)_{-3}$ and now it seems easy to solve by summation by parts:
$$\sum (2k+1)(k)_{-3}\delta k=(2k+1)\frac{(k)_{-2}}{-2}+\sum(k+1)_{-2}=\frac{2k+1}{-2(k+2)_2}-\frac{1}{k+2}=\frac{4k+3}{-2(k+2)_2}$$
And taking limits we have that the series converges to $\sum_{k\ge 0}\frac{2k+1}{(k+3)_3}=\frac{3}{4}$
The partial sum will be
$$\sum_{k=0}^{n}\frac{2k+1}{(k+3)_3}=\frac{4n+3}{-2(n+2)_2}+\frac{3}{4}$$
