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Compute the sum

\(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...\)

benjamingu22  May 29, 2017
 #1
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0

Since the denominators in your sequence follow a pattern of: 6, 24, 60, 120, 210, 336...etc. and continue at this pattern:n^3+3n^2+2n......to infinity, the reciprocal of all terms converging to 1/4. Therefore, your sequence will converge to: 2[1/4] =1/2.

Guest May 29, 2017
 #2
avatar+92781 
+1

Please can we explain your logic better guest ?

Melody  May 30, 2017
 #3
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See here: http://www.wolframalpha.com/input/?i=%E2%88%91+1%2F%5Bn%5E3%2B3n%5E2%2B2n%5D,+for+n%3D1+to+1000000

Guest May 30, 2017
 #4
avatar+92781 
0

mmm ok...

thanks for responding :)

Melody  May 30, 2017
 #5
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\(a_1=\dfrac{2}{1\cdot 2 \cdot 3}=\dfrac{1}{3}\\ a_1+a_2 = \dfrac{2}{1\cdot 2\cdot 3} + \dfrac{2}{2\cdot 3\cdot 4}=\dfrac{5}{12}\\ a_1+a_2+a_3=\dfrac{2}{1\cdot 2\cdot 3} + \dfrac{2}{2\cdot 3\cdot 4}+\dfrac{2}{3\cdot 4\cdot 5}=\dfrac{9}{20}\\ a_1 + a_2 + a_3 + a_4 = \dfrac{2}{1\cdot 2\cdot 3} + \dfrac{2}{2\cdot 3\cdot 4}+\dfrac{2}{3\cdot 4\cdot 5}+ \dfrac{2}{4\cdot 5\cdot 6}=\dfrac{7}{15}\\ a_1 + a_2 + ... + a_5 = \dfrac{7}{15}+\dfrac{2}{5\cdot 6 \cdot 7}=\dfrac{10}{21}\\ a_1 + a_2 + ... + a_6 = \dfrac{10}{21} + \dfrac{2}{6\cdot 7 \cdot 8}=\dfrac{27}{56}\)

It seems like it will not exceed 1/2.

So the answer is 1/2

MaxWong  May 30, 2017

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