13th Putnam 1953

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Problem A1

Show that (2/3) n3/2 < ∑1n √r < (2/3) n3/2 + (1/2) √n.

 

Solution

The gradient of √x is falling from x = 0 to x = n, so 2/3 n3/2 = ∫0n √x dx < √1 + √2 + √3 + ... + √n. That gives the first inequality.

To get the second, we note that the chords joining (r, √r) to (r+1, √(r+1) ) all lie under the curve, so if we subtract the area of the little triangles from ∑1n √r then we get something less than the integral. The triangles have area 1/2 √1 + 1/2 (√2 - √1) + 1/2 (√3 - √2) + ... + 1/2 (√n - √(n-1) ) = 1/2 √n. That gives the second inequality.

 


 

13th Putnam 1953

© John Scholes
jscholes@kalva.demon.co.uk
5 Mar 2002