f(n) is the number of 1s in the base 2 representation of n. Let k = ∑ f(n) / (n + n2), where the sum is taken over all positive integers. Is ek rational?
Solution
Answer: yes.
Obviously f(2n+1) = f(2n) + 1 (because the base 2 representation for 2n+1 is the same as that for 2n with an additional 1 at the end). Note also that f(2n) = f(n).
Let us look at the adjacent terms f(2n) /( 2n(2n + 1) ), and f(2n + 1) / ( (2n+1)(2n+2) ). They sum to 1/ ( (2n+1)(2n+2) ) + 2 f(2n) / ( 2n(2n+2) ) = 1/ ( (2n+1)(2n+2) ) + 1/2 f(n) / ( n(n+1) ). Thus we get k = f(1)/2 + ∑1∞ ( 1/ ( (2n+1)(2n+2) ) + 1/2 f(n) / ( n(n+1) ) = (1 - 1/2 + 1/3 - 1/4 + ... ) + 1/2 k [since 1/( (2n+1)(2n+2) ) = 1/(2n+1) - 1/(2n+2) ]. Thus k = 2 ln 2 = ln 4. So ek = 4.
© John Scholes
jscholes@kalva.demon.co.uk
16 Jan 2001