A GRE problem

Algebra Level 2

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Hana Wehbi by Hana Wehbi , 51, USA

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1 solution

Romain Bouchard
Feb 4, 2018

Relevant wiki: Telescoping Series - Sum

∑ n = 1 20 a n = ∑ n = 1 20 1 n − 1 n + 2 = ∑ n = 1 20 1 n − ∑ n = 3 22 1 n = 1 + 1 2 + ∑ n = 3 20 1 n − ( ( ∑ n = 3 20 1 n ) + 1 21 + 1 22 ) = 1 + 1 2 − ( 1 21 + 1 22 ) \sum_{n=1}^{20} a_n = \sum_{n=1}^{20} \frac{1}{n} - \frac{1}{n+2} = \sum_{n=1}^{20} \frac{1}{n} - \sum_{n=3}^{22} \frac{1}{n} = 1 + \frac{1}{2} + \sum_{n=3}^{20} \frac{1}{n} - ((\sum_{n=3}^{20} \frac{1}{n}) + \frac{1}{21} + \frac{1}{22}) = 1 + \frac{1}{2} - (\frac{1}{21} + \frac{1}{22})

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Thank you.

Hana Wehbi - 3 years, 4 months ago

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