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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

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

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  • $\begingroup$ For the second series, use $\frac{2}{x+2}-\frac{1}{x+1}$. $\endgroup$ – Jam Jun 11 '19 at 12:31
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HINT: $$\frac{1}{(x+ 1)(x+2)} = \frac{1}{x+1}-\frac{1}{x+2}$$

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Hint: Telescoping sum!

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

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  • $\begingroup$ if you still can't see if here's the spoiler: $\frac12 = 1-\frac12$; $\frac1{2*3} = \frac12-\frac13$ etc... $\endgroup$ – Book Book Book Jun 10 '19 at 19:18
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$$S_\infty =\sum_{x=0}^\infty \frac{1}{(x+ 1)(x+2)} $$ $$=\sum_{x=0}^\infty \frac{(x+2)-(x+1)}{(x+ 1)(x+2)} $$ $$=\sum_{x=0}^\infty \frac{1}{(x+1)}- \frac{1}{(x+2)} $$ 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

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  • $\begingroup$ The last line is wrong, because neither of the two constituent series converges. You are subtracting infinity from infinity. $\endgroup$ – TonyK Jun 10 '19 at 19:30
  • $\begingroup$ @TonyK Is it fine now ?? $\endgroup$ – Mr.HiggsBoson Jun 10 '19 at 19:31
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    $\begingroup$ That's better!${}$ $\endgroup$ – TonyK Jun 10 '19 at 19:46
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$$\frac12+\frac16+\frac1{12}+\frac1{20}+\frac1{30}+\cdots\to\frac12,\frac23,\frac34,\frac45,\frac56,\cdots$$

Maybe there's a pattern...

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

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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 : )

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