I have this limit: $$\lim_{n\rightarrow\infty}\sin\left(\pi\sqrt[3]{n^{3}+1}\right)$$ I don't even know if it exists. If so, what its value ? Really don't have any idea..
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1$\begingroup$ The limit of n^3 +1 as n goes to infinity is n^3. Take it from there. $\endgroup$– PhzksStdntCommented Dec 15, 2014 at 22:03
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$\begingroup$ You should be able to get a bound of the form $\sqrt[3]{n^3+1}\leq n+f(n)$ for some $f$ with $\lim_{n\to\infty}f(n)=0$. (Try factoring out $n$ from the root). What does this imply? $\endgroup$– Steven StadnickiCommented Dec 15, 2014 at 22:07
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3 Answers
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We have that $$\lim_{n\rightarrow\infty}\left(\sqrt[3]{n^3+1}-n\right)=0,$$so $\pi\sqrt[3]{n^3+1}=\pi n+ \varepsilon_n,$ where $\lim\varepsilon_n=0$. Therefore $$ \lim_{n\rightarrow\infty}\sin(\pi\sqrt[3]{n^3+1})=\lim_{n\rightarrow\infty}\sin(\pi n+\varepsilon_n)=\lim_{n\rightarrow\infty}\sin\varepsilon_n=0 $$
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3$\begingroup$ The limit does not exist. How do you know $lim sin(\pi(n))=0$, n is not necessarily an integer. $\endgroup$ Commented Oct 26, 2015 at 1:50
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$\begingroup$ Let $n$ go to infinity on the path $\frac{1}{2},\frac{5}{2},\frac{9}{2},\frac{13}{2},\frac{17}{2},...$ then $\sin(\pi n + \epsilon_n)=\cos \epsilon_n \to 1$. $\endgroup$– Math-funCommented Oct 27, 2015 at 10:47
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1$\begingroup$ To be absolutely correct it should be $\lim\limits_{n\to\infty}(-1)^n\sin\varepsilon_n=\pm 0$, but the sign does not matter since it still is $0$. $\endgroup$– zwimCommented Jan 6, 2017 at 13:38
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Hint: When $n$ is even, we have:
$$\lim_{n\rightarrow\infty}\sin\left(\pi\sqrt[3]{n^{3}+1}\right) = \lim_{n\rightarrow\infty}\sin\left(\pi\sqrt[3]{n^{3}+1} - \pi n\right) = \lim_{n\rightarrow\infty}\sin\left(\frac{\pi}{...}\right)$$
Similarly, we can deal with the odd $n$ case and conclude the limit is $0$.
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$\begingroup$ you should choose a specific convergence path so that this holds, or not? $\endgroup$– Math-funCommented Oct 27, 2015 at 10:49
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$\begingroup$ this could be related: en.wikipedia.org/wiki/Liouville_number#Irrationality_measure $\endgroup$– Math-funCommented Oct 27, 2015 at 10:50
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$\begingroup$ @Math-fun Not sure why we'd need irrationality measure here, we could simply use $\displaystyle \sqrt[3]{n^{3}+1} - n = \frac{1}{(n^3+1)^{2/3}+(n^3+1)^{1/3}n+n^2}$ which goes to $0$ as $n \to \infty$ $\endgroup$– r9mCommented Oct 27, 2015 at 11:07
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$\begingroup$ generally we do not have $\sin (f(n))=\sin(f(n)-\pi n)$. $\endgroup$– Math-funCommented Oct 27, 2015 at 11:08
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$\begingroup$ @Math-fun ah! I see your point .. we consider the even $n$ subsequence and odd $n$ subsequence separately and show both converge to $0$. $\endgroup$– r9mCommented Oct 27, 2015 at 11:13
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HINT: $\pi\sqrt[3]{n^3+1}$ is continuous and $\lim_{x\to n\pi}\sin{x}=0$.
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$\begingroup$ I don't see how the second limit applies , if $n$ is not necessarily an integer, may you please explain. @Kola B. $\endgroup$ Commented Oct 26, 2015 at 1:33