$|a_{n}| \leq C e^{-|n|} \implies \sum_{n\in \mathbb Z} a_{n} e^{in(x+iy)} $ converges absolutely for $|y|<1$?

Question

Suppose $\{a_{n}\} \subset \mathbb C$ with $|a_{n}| \leq C e^{-|n|}, n\in \mathbb Z$ and fix $C >0.$

My Question is: How to show the series, $$\sum_{n\in \mathbb Z} a_{n} e^{in (x+iy)}; (x, y \in \mathbb R)$$ converges absolutely for $|y|<1$ ?

Answer 1

Indeed we have $|a_ne^{in(x+iy)}|\leq Ce^{-|n|}e^{-ny}=Ce^{-|n|-ny}$. If we look at positive values of $n$ this is just $Ce^{-(1+y)n}$ wich is a geometric series. Thus it converges when $|e^{-(1+y)}|=e^{-(1+y)}<1$ $\Rightarrow-(1+y)<0\Rightarrow -1<y$

Next we look at negative values of $n$. Substitute $n=-m$ with $m>0$. In this case $Ce^{-|n|}e^{-ny}=Ce^{-|-m|}e^{my}=Ce^{-|m|+my}=Ce^{-m+my}=Ce^{(-1+y)m}$ wich is again a geometric series. Thus it converges when $|e^{(-1+y)}|=e^{(-1+y)}<1\Rightarrow(-1+y)<0\Rightarrow y<1$

This means that the entire series converges absolutely only when $-1<y<1$ meaning $ |y|<1$.

N.B. This will also help you to chek that it won't converge for $|y|\geq1$.