I said it does not exist because $\sin(n^2\pi/2)$ oscillates back and forth as $n$ approaches infinity. Am I correct?
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4Yes, although whether or not that's rigorous enough depends on what class it's for – Davis Yoshida Nov 19 '20 at 20:16
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5Hint : Consider the case $n$ even and the case $n$ odd. – TheSilverDoe Nov 19 '20 at 20:17
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@TheSilverDoe is correct, but maybe you are thinking too much for this very question and really want an answer for $\lim_{x\to\infty}\sin(x^2\pi/2)$? With constrictions on $x$ as part of the question? – Gyro Gearloose Nov 19 '20 at 20:33
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As suggested in the comments, we can construct two subsequences with two different limits. Let $s_n:=\sin(n^2 \pi/2)$, $a_n:=s_{2n}$, $b_{n}:=s_{2n+1}$ for all $n\geq 0$. Then $a_n=0$ for all $n$, and hence $a_n\to0$, while $b_n=1$ for all $n$, and hence $b_n\to1$. Therefore, there are two convergent subsequences with different limit, and thus the limit for $s_n$ does not exist.
On the why the limit does not exist, see here.
Monte Carlo
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