10th Putnam 1950

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Problem A3

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

 

Solution

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