An arithmotheoretic sum

For any positive integer n , let \langle n \rangle denote the closest integer to \sqrt{n}. Evaluate:

\displaystyle \mathcal{S}= \sum_{n=1}^{\infty} \frac{2^{\langle n \rangle} +2^{-\langle n \rangle}}{2^n}

(Putnam 2001)

Solution

We begin by the simple observations that \displaystyle \left ( k - \frac{1}{2} \right )^2 =k^2 - k + \frac{1}{4} and \displaystyle \left ( k + \frac{1}{2} \right )^2 =k^2 + k + \frac{1}{4}  . Combining this we get that

\langle n \rangle = k \Leftrightarrow k^2 - k + 1 < n < k^2 + k

Hence

\begin{aligned} \sum_{n=1}^{\infty} \frac{2^{\langle n \rangle} + 2^{-\langle n \rangle}}{2^n} &= \sum_{k=1}^{\infty} \sum_{\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} \left ( 2^k + 2^{-k} \right ) \left ( 2^{-k^2+k} -2^{-k^2-k} \right )\\ &= \sum_{k=1}^{\infty} \left [ 2^{-k(k-2)} - 2^{-k(k+2)} \right ]\\ &= 3 \end{aligned}

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