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How can I go about showing that $$ \sup_{t > 0} \sum_{j \in \mathbb{Z}} (t 2^{2j})^s e^{- t 2^{2j}} < \infty, \quad s > 0 $$ It doesn't seem like we can move the $\sup$ anywhere besides outside as the terms would be blowing up near $j = +\infty$. So, equivalently, how can I calculate the sum $$ \sum_{j \in \mathbb{Z}} (t 2^{2j})^s e^{- t 2^{2j}} $$ given $ t, s > 0. $

I started with the non-negative indices and viewed the sum as the $s$-derivative of $ \sum_{j \in \mathbb{Z}} e^{- t 2^{2j}}$ to evaluate which I used FTC to get an integral but it reduces to the same quantity. Of course, there should be a way to utilize the $\Gamma$-function but I am unable to spot it.

newbie
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1 Answers1

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Since $\lim\limits_{u\to \infty} u^{s+1} e^{-u} = 0$, you can find a constant $M$ such that $u^{s}e^{-u} \le \frac{M}{u}$ for $u \ge 1$.

Now fix $t>0$ and let $k\in \mathbb Z$ such that $1 \le 2^{2k}t < 2$ (it always exists)

\begin{align} \sum_{j\in \mathbb Z} \left(t2^{2j}\right)^s e^{-2^{2j}t} & \le \sum_{j \ge k} \frac{M}{2^{2j} t} + \sum_{j < k} \left(t2^{2j}\right)^s\\ &= \frac{M}{t} \frac{\frac1{2^{2k}}}{1 - \frac1{2^2}} + t^s \frac{2^{2ks}}{1 - \frac1{2^{2s}}}\\ &= \frac43 M \frac1{2^{2k} t} + \frac{2^{2s}}{2^{2s} - 1} \left(2^{2k} t\right)^s \\ &= \frac43 M + \frac{2^{2s+1}}{2^{2s} - 1} < \infty \end{align}

If you want a value of $M = (s+1)^{s+1} e^{-(s+1)}$

Kroki
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