Year: 2001
Topic: Analysis Calculus Number Theory

Problem:
For any positive integer $n$, let $\langle n\rangle$ denote the closest integer to $\sqrt{n}$.  Evaluate \[\sum_{n=1}^\infty \frac{2^{\langle n\rangle}+2^{-\langle n\rangle}}                          {2^n}.\]

Solution:
Since $(k-1/2)^2 = k^2-k+1/4$ and $(k+1/2)^2 = k^2+k+1/4$, we have that $\langle n \rangle = k$ if and only if $k^2-k+1 \leq n \leq k^2+k$.  Hence \begin{align*} \sum_{n=1}^\infty \frac{2^{\langle n \rangle} + 2^{-\langle n \rangle}}{2^n} &= \sum_{k=1}^\infty \sum_{n, \langle n \rangle = k}     \frac{2^{\langle n \rangle} + 2^{-\langle n \rangle}}{2^n} \\ &= \sum_{k=1}^\infty \sum_{n=k^2-k+1}^{k^2+k} \frac{2^k+2^{-k}}{2^n} \\ &= \sum_{k=1}^\infty (2^k+2^{-k})(2^{-k^2+k}-2^{-k^2-k}) \\ &= \sum_{k=1}^\infty (2^{-k(k-2)} - 2^{-k(k+2)}) \\ &= \sum_{k=1}^\infty 2^{-k(k-2)} - \sum_{k=3}^\infty 2^{-k(k-2)} \\ &= \boxed{3}. \end{align*}