Formula for $\sum_{k=1}^n \frac{1}{k(k+1)(k+2)}$? I was trying to find a formula for the series 
$$\sum_{k=1}^n \frac{1}{k(k+1)(k+2)} =? $$
I tried to break this into partial fractions...To see if I could telescope this series..
The partial fraction went like this 
$$\frac{1}{k(k+1)(k+2)}=\frac{1}{2k}-\frac{1}{k+1}+\frac{1}{2k+4}$$
But the terms are so random that they hardly cancel.... I also tried partial sums but couldn't make much headway....Any help to solve this would be appreciated
 A: $$\dfrac2{k(k+1)(k+2)}=\dfrac{k+2-k}{k(k+1)(k+2)}=\dfrac1{k(k+1)}-\dfrac1{(k+1)(k+2)}=f(k)-f(k+1)$$
where $f(m)=\dfrac1{m(m+1)}$
A: HINT$$\sum_{k=1}^n \frac{1}{k(k+1)(k+2)} = \sum_{k=1}^n \frac{1}{2} \left(\frac{1}{k(k+1)}-\frac{1}{(k+1)(k+2)} \right)$$
Can you take it from here?
A: Generalization:
Indeed, I will show you an approach to evaluate the series:
$$\displaystyle{\sum_{k=1}^{\infty}\frac{1}{k(k+1)\cdots(k+l)}}$$
You asked for the evaluation of a series (not finite summation, from $1$ to $n$). However, in a similar manner and without the Dominated Convergence Theorem, you can evaluate the above sum when it is finite.
First we give some definitions.
We have the definition of the beta function: 
$$\displaystyle{B(x,y)=\int_{0}^{1}{(1-t)}^{x-1}t^{y-1}dt,\ x,y>0}$$
We also have the definition of the Euler $\Gamma$ function:
$$\displaystyle{\Gamma(x)=\int_{0}^{\infty}t^{x-1}e^{-t}dt,\ x>0}$$
For the latter function, with repeated integration by parts we get that:
$$\Gamma(n)=(n-1)!,\ \forall n \in \mathbb N$$
Furthermore we need the equality below, that relates the two functions.
$$B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)},\ x,y>0$$
Now we will evaluate the series, that I mentioned in the beginning of the post.
$$\displaystyle{\sum_{k=1}^{\infty}\frac{1}{k(k+1)\cdots(k+l)}=\frac{1}{l!}\sum_{k=1}^{\infty}\frac{l!(k-1)!}{(k+l)!}=\frac{1}{l!}\sum_{k=1}^{\infty}\frac{\Gamma(l+1)\Gamma(k)}{\Gamma(l+k+1)}=\frac{1}{l!}\sum_{k=1}^{\infty}B(l+1,k)=}$$
$$\displaystyle{\frac{1}{l!}\sum_{k=1}^{\infty}\int_{0}^{1}x^l{(1-x)}^{k-1}dx=\frac{1}{l!}\int_{0}^{1}x^l\sum_{k=1}^{\infty}{(1-x)}^{k-1}dx}\ (1)$$ 
In the last step, where we interchanged the order of summation and integration, we applied the Theorem of Dominated Convergence. We also pulled the $x^l$ out of the sigma notation. By the geometric series we have that:
$$\displaystyle{\sum_{k=1}^{\infty}{(1-x)}^{k-1}=\sum_{k=0}^{\infty}{(1-x)}^{k}=\frac{1}{1-(1-x)}=\frac{1}{x}}$$
We substitute into $(1)$ and we get that:
$$\displaystyle{\sum_{k=1}^{\infty}\frac{1}{k(k+1)\cdots(k+l)}=\frac{1}{l!}\int_{0}^{1}x^{l-1}dx=\frac{1}{l\cdot l!}}$$
Finally:$$\displaystyle{\sum_{k=1}^{\infty}\frac{1}{k(k+1)\cdots(k+l)}=\frac{1}{l\cdot l!}}$$
A: \begin{align*}
S&=\sum_{k=1}^n \frac{1}{k(k+1)(k+2)}=\sum_{k=1}^n\left(\frac{1}{2k}-\frac{1}{k+1}+\frac{1}{2k+4}\right)\\
&=\frac12\sum_{k=1}^n\frac1k-\sum_{k=1}^n\frac1{k+1}+\frac12\sum_{k=1}^n\frac1{k+2}\\
&=\frac12H_n-\sum_{k=2}^{n+1}\frac1{k}+\frac12\sum_{k=3}^{n+2}\frac1{k}\\
&=\frac12H_n-\left(-1+\sum_{k=1}^{n+1}\frac1{k}\right)+\frac12\left(-1-\frac12+\sum_{k=1}^{n+2}\frac1{k}\right)\\
&=\frac14+\frac12H_n-H_{n+1}+\frac12H_{n+2}\\
&=\frac{n(n+3)}{4(n+1)(n+2)}
\end{align*}
where in the second last line we used  $$H_{n+1}=H_n+\frac1{n+1}$$
