<s>[INST] 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]
 [/INST] 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!
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