MHB How Does Grouping Terms Simplify the Infinite Series in This Week's POTW?

[Ackbach]

Here is this week's POTW:

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

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[Ackbach]

Congratulations to castor28 for a correct solution to this week's POTW, which was B-3 in the 2001 Putnam Archive. castor28's solution follows:

[sp]The idea is to group the terms of the sum by value of $k=\langle n\rangle$. Let us write $T(k)$ for the sum of the terms with a fixed value of $k$.

These terms correspond to values of $n$ in the range $\lceil (k - 1/2)^2\rceil \cdots \lfloor (k+1/2)^2\rfloor$, which boils down to $k^2-k+1\leq n \leq k^2+k$; this range contains $2k$ terms.

We have:
$$ \begin{aligned}
T(k) &= (2^k + 2^{-k})\sum_{n=k^2-k+1}^{k^2+k}{2^{-n}}\\
&=(2^k + 2^{-k}) (2^{-(k^2-k+1)}+\cdots+2^{-(k^2+k)})\\
&=(2^k + 2^{-k})(2^{2k}-1)2^{-k(k+1)}
\end{aligned}
$$

After simplification, we get:
$$ T(k) = 2^{-k(k-2)} - 2^{-k(k+2)} = T_1(k) - T_2(k)$$
We see that $T_1(k+2) = T_2(k)$. This means that, if we break down the sum as $S = S_1 + S_2$, where $S_1$ contains the terms with odd $k$ and $S_2$ contains the terms with even $k$, we get two telescoping sums: $T(k) + T(k+2) = T_1(k) + T_2(k+2)$.

As $T_2(k)\rightarrow 0$ when $k\rightarrow\infty$, we have $S_1=T_1(1)=2$ and $S_2=T_1(2)=1$ giving $3$ as the value of the sum.[/sp]