Prove that $\sum\limits_{x=0}^\infty \frac{1}{(x+ 1)(x+2)} = 1$. Prove
$$\sum_{x=0}^\infty \frac{1}{(x+ 1)(x+2)} = 1.$$
I couldn't find this problem solved online and I haven't reviewed series in a long time. I thought maybe squeeze theorem could help? A related question asks to prove
$$ \sum_{x=0}^\infty \frac{x}{(x+ 1)(x+2)} = +\infty.$$

$$S_\infty =\sum_{x=0}^\infty \frac{1}{(x+ 1)(x+2)} $$
$$=\sum_{x=0}^\infty \frac{(x+2)-(x+1)}{(x+ 1)(x+2)} $$
$$=\sum_{x=0}^\infty \frac{1}{(x+1)}- \frac{1}{(x+2)} $$
Which if you will expand and cancel 
$$ S_\infty=1-  \frac{1}{2} +\frac{1}{2}-\frac{1}{3}+\frac{1}{3}.... \infty$$
$$=1$$
a few terms , you will see that except 1 all get cancelled and you are left with 1
