Define f(n) as the number of zeros in the base 3 representation of the positive integer n. For which positive real x does F(x) = x^{f(1)}/1^{3} + x^{f(2)}/2^{3} + ... + x^{f(n)}/n^{3} + ... converge?

**Solution**

Answer: x < 25.

*Moderately hard.*

To get a handle on the problem, we collect a few terms and find that F(x) = (1/1^{3} + 1/2^{3} + 1/4^{3} + 1/5^{3} + 1/7^{3} + 1/8^{3} + ... ) + x(1/3^{3} + 1/6^{3} + 1/10^{3} + 1/11^{3} + 1/12^{3} + 1/15^{3} + 1/19^{3} + 1/20^{3} + 1/21^{3} + 1/24^{3} + ... ) + x^{2}(1/9^{3} + 1/18^{3} + 1/28^{3} + 1/29^{3} + 1/30^{3} + 1/33^{3} + ... ) + x^{3}(1/27^{3} + 1/54^{3} + ... ) + ... . This does not obviously help.

Let us focus on the m-digit numbers (in base 3). There are obviously 2 possibilities for m-1 zeros (namely 10 ... 0 and 20 ... 0). For r zeros, we can place the zeros in m-1Cr ways (the leading digit must be non-zero) and then each of the remaining m-r digits can be 1 or 2, a total of 2^{m-r} m-1Cr possibilities [aCb represents the binomial coefficient a! / ( b! a-b! ) ]. The m-digit numbers have a limited range of values, so that suggests bounds for F(x).

The smallest m-digit number is 3^{m-1} and the largest is 3^{m} - 1. So we get a lower bound by replacing 1/n^{3} by 1/3^{3m} = 1/27^{m}. In fact, the lower bound for the sum over the m-digit numbers is 1/27^{m} (x^{0} 2^{m-0} m-1C0 + x^{1} 2^{m-1} m-1C1 + ... + x^{m-1} 2^{m - (m-1)} m-1Cm-1 ) = 2^{m}/27^{m} (1 + x/2)^{k-1} = 2/27 ( (x + 2)/27 )^{k-1}. Hence the lower bound for the complete sum is 2/27 ( 1 + y + y^{2} + y^{3} + ... ), where y = (x + 2)/27, which clearly diverges for y ≥ 1 or x ≥ 25. The upper bound is 27 times larger, since we replace 1/n^{3} by 1/3^{3m-3} and it converges for y < 1 or x < 25.

© John Scholes

jscholes@kalva.demon.co.uk

7 Oct 1999