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Can I simply pass the limit in

$a_n=-\sqrt{n}\cos\sqrt{n}-\sin\sqrt{n}+\cos 1+\sin 1$

and say that this sequence diverge, because $\sqrt{n}\rightarrow\infty$?

I have found the manual solution of a book which has this exercise.

Take $\alpha=\frac{1}{2}$. Does the sequence $a_n=\int_1^{\infty}\sin x^{\alpha}\,dx$ converge or diverge?

The development of its solution leads to my problem, but the manual gives only a hint. Here it goes.

Note that being $k_n$ the sequence $k_1=31$, $k_2=314$, $k_3=314$, $k_4=3141,\dots$, i.e, $k_n$ is $10^n$ times the approximation of $\pi$ with $n$ digits. Then $$\lim\sqrt{k_n^2}\cos\sqrt{k_n^2}=+\infty \ \ \ (\mbox{Verify!})$$

The same way we can construct another sequence $p_n$ of natural number such that

$$\lim\sqrt{p_n^2}\cos\sqrt{p_n^2}=-\infty \ \ \ (\mbox{Think about it!})$$

Could anyone help me with this?

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    $\begingroup$ You could do that if $n$ were a continuous variable, but since it only takes integer values, more argument is needed. $\endgroup$ – saulspatz Mar 28 at 5:44
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    $\begingroup$ $|a_n|\geq|\sqrt{n}\cos\sqrt{n}|-3$ Does this give any idea? $\endgroup$ – user159888 Mar 28 at 5:52
  • $\begingroup$ I am sorry. My question had some typing erros. I have just corrected it $\endgroup$ – Marcos Paulo Mar 28 at 6:02
  • $\begingroup$ I don't think it helps, but $\sin^2\alpha\leqslant|\sin\alpha|\;\&\;\cos^2\alpha\leqslant|\cos\alpha|$ $\endgroup$ – Croissant Mar 28 at 12:36
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We will show that the sequence defined by

$$n\cos n$$

is divergent. Now, suppose by way of contradiction that $\cos n$ converges to $0$. I present a slightly modified version of the proof here to show this is impossible. Define the sequences

$$(x_n,y_n)=(\cos n,\sin n)$$

Then

$$x_{n+1}=x_n\cos 1-y_n\sin 1$$

$$y_{n+1}=x_n \sin 1+y_n \cos 1$$

This implies

$$0=\lim_{n\to\infty}x_{n+1}=\lim_{n\to\infty}(x_n\cos 1-y_n\sin 1)=\cos 1\lim_{n\to\infty}x_n -\sin 1\lim_{n\to\infty}y_n=-\sin 1\lim_{n\to\infty}y_n$$

$$\Rightarrow \lim_{n\to\infty}y_n=0$$

However, this leads to

$$1=\lim_{n\to\infty} 1=\lim_{n\to\infty}(x_n^2+y_n^2)=\lim_{n\to\infty}x_n^2+\lim_{n\to\infty}y_n^2=0+0=0$$

which is a contradiction. Now, since $\cos n$ does not approach $0$, there exists some subsequence of the natural numbers (let us call it $b_k$) and constant $\epsilon>0$ such that $|\cos b_k|>\epsilon$ for all $k\in\mathbb{N}$. This implies that

$$|\sqrt{n}\cos\sqrt{n}|$$

evaluated at $b_k^2$ is

$$\left|\sqrt{b_k^2}\cos\sqrt{b_k^2}\right|=|b_k\cos b_k|>b_k\epsilon$$

which clearly goes to infinity as $k$ goes to infinity. Since every other term in the sequence is bounded, and $\left|\sqrt{b_k^2}\cos\sqrt{b_k^2}\right|$ has an unbounded magnitude, we conclude

$$-\sqrt{n}\cos\sqrt{n}-\sin\sqrt{n}+\cos 1+\sin 1$$

diverges.

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  • $\begingroup$ 2 doubts: (1) Why did you leave $\cos1 \lim x_n$ out? (2) since you've proved that $\sqrt{n}\cos\sqrt{n}$ diverges, is it not necessary to worry about the convergence of $\sin\sqrt{n}$ anymore? $\endgroup$ – Marcos Paulo Mar 28 at 13:48
  • $\begingroup$ (1) We assume that $\lim x_n=0$ (by way of contradiction). (2) I proved that a subsequence of $\sqrt{n}\cos\sqrt{n}$ diverges to either positive or negative infinity. Since $\sin\sqrt{n}$ is bounded, it doesn't effect the divergence of $\sqrt{n}\cos\sqrt{n}$ $\endgroup$ – QC_QAOA Mar 28 at 15:26

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