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
 A: 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:
$$S_N = -\frac52\left(H_N + \frac{1}{N+1} + \frac{1}{N+2}-\frac32\right) + 3\left(H_N+\frac{1}{N+1}-1\right) - \frac12H_N=\\=
\frac34 + \frac{1}{2(N+1)}-\frac{5}{2(N+2)}.$$
A: 
The good old way: compute a few terms and find a pattern.
$$\frac2{12},\frac7{24},\frac{15}{40},\frac{26}{60},\frac{40}{84}\cdots$$
By finite differences, we find that the numerators are $\dfrac{n(3n+1)}2$ and the denominators $2(n+1)(n+2)$.

Check:
$$S_1=\frac2{12}$$and
$$S_n-S_{n-1}=\frac{n(3n-1)}{4(n+1)(n+2)}-\frac{(n-1)(3n-4)}{4n(n+1)}=\frac{2n-1}{n(n+1)(n+2)}=a_n.$$
A: 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
$$\begin{align}\sum_{i=1}^{n}\frac{2i-1}{i(i+1)(i+2)}&=\sum_{i=1}^{n}\left(\frac{2i-\frac 12}{i(i+1)}-\frac{2(i+1)-\frac 12}{(i+1)(i+2)}\right)\\&=\frac{2\cdot 1-\frac 12}{1\cdot (1+1)}-\frac{2(n+1)-\frac 12}{(n+1)(n+2)}\\&=\color{red}{\frac 34-\frac{4n+3}{2(n+1)(n+2)}}\end{align}$$
A: 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
$$\sum_{k=0}^{n}\frac{2k+1}{(k+3)_3}=\frac{4n+3}{-2(n+2)_2}+\frac{3}{4}$$
A: see that $a_n = (2n-1)/((n*(n+1)*(n+2)=-1/2*(1/n)+3*(1/(n+1))-5/2*(1/(n+2)) = 1/2* (-1/n+1/(n+1)+5*(1/(n+1)-1/(n+2))$, so $S_n = 1/2*(-1+5+1/(n+1)-5/(n+2)) = 1/2*(4n^2+8n+5)/((n+1)*(n+2))$.
