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I need to calculate sequence limit of $\left\{a_n\right\}_{n=1}^{\infty}$ where $\left\{a_n\right\}_{n=1}^{\infty} = \left|\sin \left(\pi \cdot \sqrt[3]{n^3+n^2}\right)\right|$

$$ \begin{aligned} & \quad \lim _{n \rightarrow \infty}\left\{a_n\right\}_{n=1}^{\infty}=\lim _{n \rightarrow \infty}\left|\sin \left(\pi \cdot \sqrt[3]{n^3+n^2}\right)\right|= \\ & =\lim _{n \rightarrow \infty}\left|\sin \left(\pi \cdot n \cdot \sqrt[3]{1+\frac{1}{n}}\right)\right| \end{aligned} $$

I then assume that since infinitely large natural $n$ make $1 + \frac{1}{n}$ almost 1, the $\sin$ argument is almost $\pi n$ so the $\sin$ value hovers around $0$. Which is my answer, which is incorrect. Where is the flaw in my logic? Thank you in advance

Alex Silver
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  • Wolfram Alpha says it's indeterminate.
    Try evaluating Limit[|sin(pi*(n^3+n^2)^1/3)|,n->∞]

    https://www.wolframalpha.com/input?i2d=true&i=Limit%5B%7Csin%5C%2840%29pi*Power%5B%5C%2840%29Power%5Bn%2C3%5D%2BPower%5Bn%2C2%5D%5C%2841%29%2CDivide%5B1%2C3%5D%5D%5C%2841%29%7C%2Cn-%3E%E2%88%9E%5D

     – nickalh Apr 13 '24 at 15:16
  • @nickalh Thank you for your comment. The proposed solution is: $a_n=\left|\sin \left(\pi \sqrt[3]{n^3+n^2}-\pi n\right)\right|$

    $$ f(x)=\pi \sqrt[3]{x^3+x^2}-\pi x=\pi \frac{x^2}{\left(x^3+x^2\right)^{\frac{2}{3}}+x \sqrt[3]{x^3+x^2}+x^2} \underset{x \rightarrow+\infty}{\longrightarrow} \frac{\pi}{3} $$

    $$ \lim _{x \rightarrow+\infty}|\sin (f(x))|=\sin \left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}, $$

    I have seen Wolfram's take on this, and I don't understand how it correltes to the proposed solution (even though I understand it).

     – Alex Silver Apr 13 '24 at 15:19
  • This might help. – mathlove Apr 13 '24 at 15:33
  • That's a different sequence. You're subtracting $\pi n$. I would expect totally different results to a different sequence. Is this a step in proving it based on something derived from $0<|n-m|<\delta$? – nickalh Apr 13 '24 at 15:33
  • @nickalh Thank you. The proposed solution states that $$ \lim {n \rightarrow \infty}\left{a_n\right}{n=1}^{\infty}=\lim _{n \rightarrow \infty}\left|\sin \left(\pi \cdot \sqrt[3]{n^3+n^2}\right)\right|= \lim _{n \rightarrow \infty}\left|\sin \left(\pi \cdot \sqrt[3]{n^3+n^2} - \pi n\right)\right| \ $$ as the function period of function $|\sin x|$ is $\pi$, so (I'm assuming this part), we can add or subtract any number of $\pi$s and the $|\sin|$ wouldn't change – Alex Silver Apr 13 '24 at 16:19

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For $b_n$ which satisfies $b_n \to 1$, $a_n$ be $$a_n = \left|\sin(\pi n b_n)\right|$$

Your attempt is Since $b_n \to 1$, $\pi nb_n$ may be approximated to $$\pi nb_n \approx \pi n$$

which is False.

You can check above easily with these two examples : $$\begin{align} & b_n = 1 + \frac{1}{2n} \\ & b_n = 1+\frac{1}{3n}\end{align}$$

Solution

We can solve this problem with two ways : the first one is, as you noticed, subtract $n\pi$ and use periodicity.

Another way with using taylor expansion : for large $n$,

$$\sqrt[3]{1+\frac{1}{n}}\approx1+\frac{1}{3n}$$

So your limit will be $$\begin{align} a_n &\approx \left|\sin\left(n\pi\left(1+\frac{1}{3n}\right)\right)\right| \\ &\approx \left|\sin\left(n\pi + \frac{\pi}{3}\right)\right| \\ &\approx \left|\sin\left(\frac{\pi}{3}\right)\right| \\ &= \frac{\sqrt{3}}{2}.\end{align}$$

bFur4list
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