Let @n denote the closest integer to √n. Find ∑ (2@n + 2-@n)/2n, 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 = n2 + 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/2n and ∑ 2-@n/2n separately. The first is (20 + 2-1) + (2-1 + 2-2 + 2-3 + 2-4) + (2-4 + ... + 2-9) + ... = (1 + 1/2 + 1/4 + ... ) + (1/2 + 1/24 + 1/29 + 1/216 + ... ).
The second sum is (1/22 + 1/23) + (1/25 + ... + 1/28) + (1/210 + ... + 1/215) + ... .
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