20th Putnam 1959

Problem B2

Show that any positive real can be expressed in infinitely many ways as a sum ∑ 1/(10 an), where a1 < a2 < a3 < ... are positive integers.



∑ 1/n diverges, so ∑ 1/(10n) diverges. Let k be any positive real. Take N such that 1/(10N) < k. Let M be any integer > N. We show how to find an expression k = ∑ 1/(10an) with a1 = M. Take enough terms a2 = 1/(10(M+1)), a3 = 1/(10(M+2)), ... so that ∑ 1/(10an) < k, but adding another term would give a sum ≥ k. Let the difference k - ∑ 1/(10an) be k1 > 0. Now take M1 so that 1/(10 M1) < k1 and repeat. [In other words, take as many terms 1/(10 M1) + 1/(10(M1+1)) + ... as we can whilst keeping the sum < k1.] Let k2 be the new difference, and so on. This process gives a sum which converges to k. Each such sum is different because it has a different starting term.



20th Putnam 1959

© John Scholes
15 Feb 2002