Without direct evaluation, find the minimum integer value of r in this expression :
1+1/√2+1/√3+......+1/√2018 ≤ r*√2018
Note: Adapted from a problem which appeared at Round 1 of Singapore Mathematical Olympiad Open, 2018.
The rhs is less than the integral from 0 to 2018 of 1/sqrt(x) dx which equals 2*sqrt(2018)
2*sqrt(2018) <= r*sqrt(2018)
implies r is at least 2
Can r=1?
No.
the rhs is greater than the integral from 1 to 2019 of 1/sqrt(x) dx which equals 2*sqrt(2019)-1
(2*sqrt(2019)-1)/sqrt(2018) is clearly greater than 1.
So r=2
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Posted by Jer
on 2025-02-20 13:32:03 |
calculus solution
