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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 k=12k(k+1)(k+2).

We can perform partial fractions to get a telescoping sum.

 

Note that k=1(f(k)f(k+1))=f(1)limkf(k) for any function f.

 

Let 2k(k+1)(k+2)=Ak+Bk+1+Ck+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 {2A=2B=22C=2.

 

Then A = C = 1, B = -2.

 

We have:

k=12k(k+1)(k+2)=k=1(1k+1k+22k+1)=k=1(1k1k+1)k=1(1k+11k+2)

 

Can you continue with my hints here?

 May 3, 2022
edited by MaxWong  May 3, 2022

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