Let @n denote the closest integer to √n. Find ∑ (2^{@n} + 2^{-@n})/2^{n}, where the sum is taken from n = 1 to n = infinity.

**Solution**

Answer: 3.

We need to find a more useful expression for @n. We have (n + 1/2)^{2} = n^{2} + n + 1/4, which lies between the integers n(n+1) and n(n+1)+1. Thus the 2n values @m for m = (n-1)n+1, (n-1)n+2, ... , (n-1)n+2n=n(n+1) are all n.

The sum is obviously absolutely convergent, so we can consider ∑ 2^{@n}/2^{n} and ∑ 2^{-@n}/2^{n} separately. The first is (2^{0} + 2^{-1}) + (2^{-1} + 2^{-2} + 2^{-3} + 2^{-4}) + (2^{-4} + ... + 2^{-9}) + ... = (1 + 1/2 + 1/4 + ... ) + (1/2 + 1/2^{4} + 1/2^{9} + 1/2^{16} + ... ).

The second sum is (1/2^{2} + 1/2^{3}) + (1/2^{5} + ... + 1/2^{8}) + (1/2^{10} + ... + 1/2^{15}) + ... .

Now (1 + 1/2 + 1/4 + ... ) = 2, and the other two brackets together give (1/2 + 1/4 + 1/8 + ... ) = 1.

© John Scholes

jscholes@kalva.demon.co.uk

16 Dec 2001