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Compute the sum     
2/(1•2•3)+2/(2•3•4)+2/(3•4•5)+•••

 May 3, 2022
 #1
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sumfor(n, 1, infinity, (2 / (n*(n+1)*(n+2)))==It converges to 1/2

 May 3, 2022
 #2
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The sum, written in sigma notation, is \(\newcommand{\dsum}{\displaystyle\sum} \dsum_{k = 1}^{\infty} \dfrac2{k(k + 1)(k + 2)}\).

We can perform partial fractions to get a telescoping sum.

 

Note that \(\newcommand{\dsum}{\displaystyle\sum} \dsum_{k = 1}^\infty (f(k) - f(k + 1)) = f(1) - \lim_{k\to\infty} f(k)\) for any function f.

 

Let \(\dfrac2{k(k + 1)(k + 2)} = \dfrac A{k} + \dfrac B{k + 1} + \dfrac C{k + 2}\) for some constants A, B, C.

 

Multiplying both sides by k(k + 1)(k + 2) gives \(2 = A(k + 1)(k + 2) + Bk(k + 2) + Ck(k + 1)\).

 

Substituting k = 0, k = -1, and k = -2 respectively gives \(\begin{cases}2A = 2\\-B = 2\\2C = 2\end{cases}\).

 

Then A = C = 1, B = -2.

 

We have:

\(\newcommand{\dsum}{\displaystyle\sum} \begin{array}{rcl} \dsum_{k = 1}^{\infty} \dfrac2{k(k + 1)(k + 2)} &=& \dsum_{k = 1}^\infty \left(\dfrac1k + \dfrac1{k + 2} - \dfrac2{k + 1}\right)\\ &=& \dsum_{k = 1}^\infty \left(\dfrac1k - \dfrac1{k + 1}\right) - \dsum_{k = 1} ^\infty\left( \dfrac1{k + 1} - \dfrac1{k + 2}\right) \end{array}\)

 

Can you continue with my hints here?

 May 3, 2022
edited by MaxWong  May 3, 2022

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