Prove that $\lim\limits_{n\to\infty} e^{i n t}$ does not exist if $0

Question

How does one prove that? Surely its not that hard to prove and it seems to me I have seen a solution somewhere but now I can not recall where it was.

Consider $\left|,e^{int}-e^{i(n+1)t},\right|=\left|,1-e^{it},\right|$ and show that this prevents the sequence from being Cauchy. — robjohn, Oct 03 '18 at 18:34

Larry B. · Accepted Answer · 2018-10-03T18:14:58.097

3

I based this proof on the answer to a similar question.

Suppose $\lim_{n\to\infty} e^{it\cdot n} = x$ as the limit. Then $x = \lim_{n\to\infty} e^{it\cdot (n + 1)} = e^{it}\lim_{n\to\infty} e^{itn} = e^{it}x$ But that would mean $e^{it} = 1$ or $x=0$. The case where $e^{it} = 1$ is impossible with the bounds on $t$. And $x\neq 0$ since $|e^{iv}| = 1$ for any value of $v$, meaning the limit must also lie on that circle. (Thank you, Andrei, for mentioning the $x=0$ case.

edited Oct 03 '18 at 18:14

answered Oct 03 '18 at 18:09

Larry B.

3,384

1

Unless $x=0$. You need to check this case as well (trivial). – Andrei Oct 03 '18 at 18:13
Edited, thank you @Andrei – Larry B. Oct 03 '18 at 18:15
That's it! Thank you. – 9Rx4FHm Oct 03 '18 at 18:39

Prove that $\lim\limits_{n\to\infty} e^{i n t}$ does not exist if $0

1 Answers1