telescopic series & sum of sequences - how to? The entire web shows the same example basicly(or something extremely similar): $$\sum_{n=2}^\infty \frac{-2}{(n+1)(n+2)}= \frac{-2}{3}$$
Is there a way to find this final answer without doing the pretty long procedure?    
In addition to it I want to apply the telescopic series principal on the following sequence: $$(a_{n+1} - a_n)$$
How would I do it? [$a_n$ is sequence and $a_{n+1} - a_n$ is bounded]
 A: Hint: Telescoping 
Notice that 
$$\frac{-2}{(n+1)(n+2)}=-2\left(\frac{1}{n+1}-\frac{1}{n+2}\right)$$
See Partial fraction Decomposition on Wikipedia or on Mathworld
I hope you can take it from here
Formula for Partial sum:
$$a_{p}=\sum_{n=0}^p\frac{-2}{(n+1)(n+2)}=\frac{-2(p+1)}{p+2}$$
A: If the sequence $a_n$ converges to $a$, then
$$\sum_{n=1}^\infty(a_{n+1}-a_n)\stackrel1=\lim_{n\to\infty}\sum_{k=1}^n(a_{k+1}-a_k)\stackrel2=\lim_{n\to\infty}(a_{n+1}-a_1)\stackrel3=a-a_1$$
Step $2$:
$$\sum_{k=1}^n(a_{k+1}-a_k)=(a_2-a_1)+(a_3-a_2)+\ldots+(a_{n+1}-a_n)$$
But you can cancel all the terms but $a_{n+1}$ and $a_1$, so the sum is $a_{n+1}-a_1$.
A: Yeah, so basically the only sure way of completing this series is by doing the partial decomposition of the original series. Then by finding the coefficients of the fractions you should be able to see a cancellation. The only step left is to find the nth terms and write out the terms the terms that don't cancel which will be your answer.  
A: By partial fractions,
$$\frac{-2}{(n+1)(n+2)} \equiv \frac{A}{n+1} + \frac{B}{n+2}$$
For some $A$ and $B$.
Thus,
$$-2 \equiv A(n+2)+B(n+1)$$
By letting $n=-1$ and $n=-2$ in turn, we can see that $A=-2$ and $B=2$. So the sum is now:
$$\sum_{n=2}^{n=k}{\left(\frac{2}{n+2} - \frac{2}{n+1}\right)}=2\sum_{n=2}^{n=k}{\left(-\frac{1}{n+1} + \frac{1}{n+2}\right)}$$
Which is
$$2\left(\left(-\frac{1}{3}+\frac{1}{4}\right)+\left(-\frac{1}{4}+\frac{1}{5}\right)+\left(-\frac{1}{5}+\frac{1}{6}\right)+...+\left(-\frac{1}{k}+\frac{1}{k+1})\right)+\left(-\frac{1}{k+1}+\frac{1}{k+2}\right)\right)$$
Which, by the method of differences, is equivalent to
$$2\left(-\frac{1}{3}+\frac{1}{k+2}\right)$$
If we limit $k$ to $\infty$, $\frac{1}{k+2}$ becomes $0$, thus our sum is $$-\frac{2}{3}$$
More generally, by the same method, $$\sum_{n=x}^{n=y}{(a_{n+1}-a_{n}})=a_{y+1}-a_{x}$$
Hope this helped!
A: A well-known generalization
of your sum is this:
Consider the product of
the $k$ consecutive integers
starting at $n$,
$
n(n+1)...(n+k-1)
$.
Then
$\begin{array}\\
\dfrac1{n(n+1)...(n+k-1)}-\dfrac1{(n+1)...(n+k)}
&=\dfrac{(n+k)-n}{n(n+1)...(n+k)}\\
&=\dfrac{k}{n(n+1)...(n+k)}\\
\end{array}
$
or
$\dfrac{1}{n(n+1)...(n+k)}
=\dfrac1{k}\left(\dfrac1{n(n+1)...(n+k-1)}-\dfrac1{(n+1)...(n+k)}\right)
$.
Adding $1$ to $k$,
$\dfrac{1}{n(n+1)...(n+k-1)}
=\dfrac1{k-1}\left(\dfrac1{n(n+1)...(n+k)}-\dfrac1{(n+1)...(n+k+1)}\right)
$.
Therefore
$\begin{array}\\
\sum_{n=1}^m \dfrac1{n(n+1)...(n+k-1)}
&=\sum_{n=1}^m \dfrac1{k-1}\left(\dfrac1{n(n+1)...(n+k)}-\dfrac1{(n+1)...(n+k+1)}\right)\\
&=\dfrac1{k-1}\sum_{n=1}^m \left(\dfrac1{n(n+1)...(n+k)}-\dfrac1{(n+1)...(n+k+1)}\right)\\
&=\dfrac1{k-1}\left(\dfrac1{1\cdot 2 ...\cdot (k+1)}-\dfrac1{(m+1)...(m+k+1)}\right)\\
&=\dfrac1{k-1}\left(\dfrac1{(k+1)!}-\dfrac1{(m+1)...(m+k+1)}\right)\\
\end{array}
$
Letting $m \to \infty$,
$\sum_{n=1}^{\infty} \dfrac1{n(n+1)...(n+k-1)}
=\dfrac1{(k-1)(k+1)!}
$.
