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I have the following problem:

For any positive integer n, let $\langle n \rangle$ denote the integer nearest to $\sqrt n$.

(a) Given a positive integer $k$, describe all positive integers $n$ such that $\langle n \rangle = k$.

(b) Show that $$\sum_{n=1}^\infty{\frac{2^{\langle n \rangle}+2^{-\langle n \rangle}}{2^n}}=3$$

My progress: The first one is rather easy. As $$\left( k-\frac{1}{2} \right) < \sqrt n < \left( k+\frac{1}{2} \right) \implies \left( k-\frac{1}{2} \right)^2 < n < \left( k+\frac{1}{2} \right)^2 \implies \left( k^2-k+1 \right) \leq n \leq \left( k^2+k \right)$$ Actually, there would be $2k$ such integers.

But, I have no idea how to approach the second problem. Please give me some hints.

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  • $\begingroup$ If you can show that $\frac{2^{\langle n \rangle}+2^{-\langle n \rangle}}{2^n}=\left(\frac{3}{4}\right)^n$ then this series is a geometric series with $x=\frac34$ that starts at $1$ and therefore is equal to $3$. $\endgroup$ – Jimmy R. Dec 4 '15 at 11:28
  • $\begingroup$ @Stef You're right... Let's see if anything can be done by using it... $\endgroup$ – SinTan1729 Dec 4 '15 at 11:32
  • $\begingroup$ I tried but nothing so far... I do not think that this is true, (by writting down the first few terms) so do not waste too much time on this. Ah, and of course you should use somewhere question (a). $\endgroup$ – Jimmy R. Dec 4 '15 at 11:32
  • $\begingroup$ @Stef But, for $n=1$ the value is $\frac{3}{4}$ $\endgroup$ – SinTan1729 Dec 4 '15 at 11:39
  • $\begingroup$ Yes, only for $n=1$. For $n=2, 3$ and $4$ it does not work. $\endgroup$ – Jimmy R. Dec 4 '15 at 11:41
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Direct computation seems to work. I recall this problem being an old Putnam problem.

We have $\langle n \rangle = k$, iff $n \in [k^2 - k + 1, k^2 + k]$.

Now, let's compute this sum. $$ \sum_{n = 1}^\infty \frac{2^{\langle n \rangle} + 2^{-\langle n \rangle}}{2^n} = \sum_{k=1}^\infty \sum_{i=k^2 - k + 1}^{k^2 + k} \frac{2^k + 2^{-k}}{2^i} = \sum_{k=1}^\infty \frac{2^{2k}+1}{2^k}\sum_{i=k^2 - k + 1}^{k^2 + k} \frac1{2^i} = \sum_{k=1}^\infty \frac{2^{2k}+1}{2^{k^2+1}}\sum_{i=0}^{2k-1} \frac{1}{2^i} = \sum_{k=1}^\infty \frac{2^{2k}+1}{2^{k^2 + 1}}\frac{1 - \frac{1}{4^k}}{1-\frac{1}{2}} = \sum_{k=1}^\infty \frac{2^{2k}+1}{2^{k^2 + 1}}\frac{2^{2k}-1}{2^{2k-1}} = \sum_{k=1}^\infty \frac{2^{4k}-1}{2^{k^2 + 2k}} = \sum_{k=1}^\infty \left(2^{1-(k-1)^2} - 2^{1-(k+1)^2}\right) = 2^1 + 2^0 = 3$$

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  • $\begingroup$ Wow... This was great! $\endgroup$ – SinTan1729 Dec 4 '15 at 11:53
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The idea is to rewrite the sum as a double sum by observing that $$\langle m^2 + k \rangle = m$$ for $k \in \{-m+1, \ldots, m\}$. Therefore, $$\begin{align*} S &= \sum_{n=1}^\infty \frac{2^{\langle n \rangle} + 2^{-\langle n \rangle}}{2^n} \\ &= \sum_{m=1}^\infty \sum_{k=-m+1}^{m} \frac{2^m + 2^{-m}}{2^{m^2+k}} \\ &= \sum_{m=1}^\infty \frac{2^m + 2^{-m}}{2^{m^2}} \sum_{k=-m+1}^m \frac{1}{2^k} \\ &= \sum_{m=1}^\infty \frac{2^m + 2^{-m}}{2^{m^2}} \left(2^m - 2^{-m}\right) \\ &= \sum_{m=1}^\infty \frac{2^{2m} - 2^{-2m}}{2^{m^2}} \\ &= \sum_{m=1}^\infty 2^{-m(m-2)} - 2^{-m(m+2)} \\ &= \sum_{m=1}^\infty 2^{-m(m-2)} - \sum_{k=3}^\infty 2^{-(k-2)k} \\ &= \sum_{m=1}^2 2^{-m(m-2)} = 2^1 + 2^0 = 3.\end{align*}$$

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