10th Putnam 1950

The sequence a_n is defined by a₀ = α, a₁ = β, a_n+1 = a_n + (a_n-1 - a_n)/(2n). Find lim a_n.

Solution

a_n+1 - a_n = -1/2n (a_n - a_n-1), hence a_n+1 - a_n = (-1)ⁿ/(2·4·6 ... .2n) (a₁ - a₀). Then a_n = (a_n - a_n-1) + (a_n-1 - a_n-2) + ... + (a₁ - a₀) + a₀.

Hence the limit is β + (β - α)/2 - (β - α)/2·4 + (β - α)/2·4·6 - ... . But 1/2 -1/2·4 + 1/2·4·6 - ... = 1 - 1/√e, so the limit is α + (β - α)/√e.

10th Putnam 1950

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