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
$$ \sum_{x=0}^\infty \frac{x}{(x+ 1)(x+2)} = +\infty.$$
 A: Hint: Telescoping sum! 
$$1-\frac12+\frac12-\frac13+\frac13-\frac14+...... = 1$$
A: $$S_\infty =\sum_{x=0}^\infty \frac{1}{(x+ 1)(x+2)} $$
$$=\sum_{x=0}^\infty \frac{(x+2)-(x+1)}{(x+ 1)(x+2)} $$
$$=\sum_{x=0}^\infty \frac{1}{(x+1)}- \frac{1}{(x+2)} $$
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
A: HINT: $$\frac{1}{(x+ 1)(x+2)} = \frac{1}{x+1}-\frac{1}{x+2}$$
A: $$\frac12+\frac16+\frac1{12}+\frac1{20}+\frac1{30}+\cdots\to\frac12,\frac23,\frac34,\frac45,\frac56,\cdots$$
Maybe there's a pattern...
A: $$-\log(1-x)=\sum_{k=0}^{\infty}\frac{x^{k+1}}{k+1}$$
$$-\int_{0}^{1}\log(1-x)dx=\int_{0}^{1}\sum_{k=0}^{\infty}\frac{x^{k+1}}{k+1}dx$$
$$1=\sum_{k=0}^{\infty}\frac{1}{(k+1)(k+2)}$$
A: Break the fraction 
1/(x+1)(x+2)
 into two partial fractions as
 A/(x+1) and
 B/(x+2). 
Now equate the first fraction with sum of two partial fractions. You get
 A(x+2) + B(x+1) = 1
Put x=-1and then -2 and obtain the value of A and B. 
Put the sum of the resolved two partial fractions into the original sum. You get 
sum [1/(x+1)]-sum [1/(x+2)]
 and open the terms.
You get a expansion of 
S= 1 + 1/2 + 1/3+ ••• -1/2 - 1/3 - ••• 
So see all cancel and only one remains
Hence we have the sum as 1
: )
