Infinite sum?

Algebra Level 2

Find the value of ( 1 1 + 1 1 + 2 + 1 1 + 2 + 3 + 1 1 + 2 + 3 + 4 + 1 1 + 2 + 3 + 4 + 5 + … ) 2 \small \left(\frac { 1 }{ 1 } +\frac { 1 }{ 1+2 } +\frac { 1 }{ 1+2+3 } +\frac { 1 }{ 1+2+3+4 } +\frac { 1 }{ 1+2+3+4+5 } +\ldots\right)^{ 2 }


The answer is 4.

Supawish Kanokpongsakorn by Supawish Kanokpongsakorn , 28, Thailand

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1 solution

( 1 1 + 1 1 + 2 + 1 1 + 2 + 3 + 1 1 + 2 + 3 + 4 + . . . ) 2 = ( 1 1 + 1 3 + 1 6 + 1 10 + . . . ) 2 = [ 1 1 + 2 ( 1 6 + 1 12 + 1 20 + . . . ) ] 2 = [ 1 + 2 ( 1 2 ⋅ 3 + 1 3 ⋅ 4 + 1 4 ⋅ 5 + . . . ) ] 2 = [ 1 + 2 ( 1 2 − 1 3 + 1 3 − 1 4 + 1 4 − 1 5 + . . . ) ] 2 = [ 1 + 2 ( 1 2 ) ] 2 = ( 1 + 1 ) 2 = 4 (\frac { 1 }{ 1 } +\frac { 1 }{ 1+2 } +\frac { 1 }{ 1+2+3 } +\frac { 1 }{ 1+2+3+4 } +...)^{ 2 }\quad =\quad (\frac { 1 }{ 1 } +\frac { 1 }{ 3 } +\frac { 1 }{ 6 } +\frac { 1 }{ 10 } +...)^{ { 2 } }\\ =\quad [\frac { 1 }{ 1 } +2(\frac { 1 }{ 6 } +\frac { 1 }{ 12 } +\frac { 1 }{ 20 } +...)]^{ { 2 } }\\ =\quad [1+2(\frac { 1 }{ 2\cdot 3 } +\frac { 1 }{ 3\cdot 4 } +\frac { 1 }{ 4\cdot 5 } +...)]^{ 2 }\\ =\quad [1+2(\frac { 1 }{ 2 } -\frac { 1 }{ 3 } +\frac { 1 }{ 3 } -\frac { 1 }{ 4 } +\frac { 1 }{ 4 } -\frac { 1 }{ 5 } +...)]^{ { 2 } }\\ =\quad [1+2(\frac { 1 }{ 2 } )]^{ { 2 } }\\ =\quad (1+1)^{ { 2 } }\\ =\quad 4

6 Helpful 0 Interesting 1 Brilliant 0 Confused

Nice solution! Alternatively, we can just evaluate the sum inside of the parentheses, and then square it.

Using the Sum of n, n², or n³ , we know that 1 + 2 + … + n = n ( n + 1 ) 2 , 1+2+\ldots+n = \frac{n(n+1)}{2}, so the sum is ∑ k = 1 ∞ 2 k ( k + 1 ) . \sum_{k=1}^{\infty} \frac{2}{k(k+1)}.

Using Partial Fractions , we can write this as ∑ k = 1 ∞ 2 k − 2 k − 1 . \sum_{k=1}^{\infty} \frac{2}{k} - \frac{2}{k-1}.

This is a Telescoping Series - Sum , and all the terms cancel except for 2 1 = 2 , \frac{2}{1} = 2, so the sum is 2. Thus, our answer is 2 2 = 4. 2^2 = 4.

Eli Ross Staff - 5 years, 7 months ago

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