Prove that $\sum\limits_{x=0}^\infty \frac{1}{(x+ 1)(x+2)} = 1$.

Question

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

$$ \sum_{x=0}^\infty \frac{x}{(x+ 1)(x+2)} = +\infty.$$

Answer 1

Hint: Telescoping sum!

$$1-\frac12+\frac12-\frac13+\frac13-\frac14+...... = 1$$

Answer 2

HINT: $$\frac{1}{(x+ 1)(x+2)} = \frac{1}{x+1}-\frac{1}{x+2}$$

Answer 3

$$S_\infty =\sum_{x=0}^\infty \frac{1}{(x+ 1)(x+2)} $$ $$=\sum_{x=0}^\infty \frac{(x+2)-(x+1)}{(x+ 1)(x+2)} $$ $$=\sum_{x=0}^\infty \frac{1}{(x+1)}- \frac{1}{(x+2)} $$ 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

Answer 4

$$\frac12+\frac16+\frac1{12}+\frac1{20}+\frac1{30}+\cdots\to\frac12,\frac23,\frac34,\frac45,\frac56,\cdots$$

Maybe there's a pattern...

Answer 5

$$-\log(1-x)=\sum_{k=0}^{\infty}\frac{x^{k+1}}{k+1}$$ $$-\int_{0}^{1}\log(1-x)dx=\int_{0}^{1}\sum_{k=0}^{\infty}\frac{x^{k+1}}{k+1}dx$$ $$1=\sum_{k=0}^{\infty}\frac{1}{(k+1)(k+2)}$$

Answer 6

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 : )