Limit superior and inferior when one part diverges

Question

How can I find the limit superior and inferior of given sequence: $x_n = (1 + \frac{1}{2n})\cos{\frac{n\pi }{3}}$ as $ n \in \mathbb N $

I did the following: since $\lim_{n\to\infty}(1 + \frac{1}{2n}) = 1$ and the second term' limit oscillates between -1 and 1, I decided that the supremum should be $1$ and infimum should be $-1$. But this is an incorrect answer.

What I do wrong? Thank you.

UPD: It is a task from the online-courses site and there is an automatic answer checking system. So, as I said in my question, answer $-1$ and $1$ not passed and there is no explanation why.

I believe there should be used Bolzano–Weierstrass theorem to find all possible convergent subsequences, and then find limits for each of them...

Answer 1

We have $\lim_{n\to\infty}(1+\frac{1}{2n}) = 1$ and $0 \le |\cos(x)| \le 1$ for all $x \in \Bbb R$, so $$\limsup_n \,(1+\frac{1}{2n}) = 1 \text{ and } \limsup_n \, \left|\cos{\frac{n\pi }{3}} \right| \le 1.$$ Using $\limsup ( a_n b_n ) \leqslant \limsup a_n \limsup b_n$ (with $a_n = 1 + \frac{1}{2n}$ and $b_n = |\cos\frac{n\pi}{3}|$), we have $$\limsup_n \, (1 + \frac{1}{2n}) \left|\cos{\frac{n\pi }{3}} \right| \le 1,$$ so $$-1 \le -\limsup_n \, (1 + \frac{1}{2n}) \left|\cos{\frac{n\pi }{3}} \right| \le \liminf_n\, (1 + \frac{1}{2n}) \cos{\frac{n\pi }{3}}.$$

(Recall that $\liminf_n a_n$ can be equivalently defined as the minimum possible limit of subsequence $(a_{n_k})_k$, which is bounded below by $(-|a_{n_k}|)_k$.)

• $n = 6k + 3$: $x_n = (1 + \frac{1}{12k + 6})(-1) \to -1$ as $k \to \infty$.
• $n = 6k$: $x_n = (1 + \frac{1}{12k})(1) \to 1$ as $k \to \infty$.

This shows that $\limsup_n x_n = 1$ and $\liminf_n x_n = -1$.

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(Edit in response to OP's update)

OP has appealed to his online system that we have no access to to conclude that my answer is wrong.

Here, I'm going to prove that why $$\limsup_n \,x_n < 1+\epsilon \quad \forall \epsilon > 0.$$ In other words, wny number $1 + \epsilon$ strictly greater than $1$ can't be a limit of any subsequence of $(x_n)_n$. With the example subsequence $n = 6k$ in the last part of my original answer, this would show that $\limsup_n x_n = 1$.

The idea is simple: $\require{cancel}$ $$x_n = \underbrace{\left( \color{red}{1+\frac{1}{2n}} \right)}_{\to 1} \; \xcancel{\color{blue}{\cos{\frac{n\pi }{3}}}}.$$

Source: Wikipedia's page on $\limsup$ in French

To make it rigorous, for any $\epsilon > 0$, take $N = \left\lceil \dfrac{1}{\epsilon} \right\rceil$. For all $n \ge N$, $$\left( 1 + \dfrac{1}{2n} \right) \cos{\frac{n\pi }{3}} \le \left( 1 + \dfrac{1}{2n} \right) \, (1) \le 1 + \dfrac12\:\epsilon < 1 + \epsilon,$$ which violates the later part of (1) in the proposition in an older question: the above inequality shows that $x_n \le 1 + \dfrac\epsilon2$ eventually, instead of $x_n > (1 + \epsilon) - \dfrac\epsilon2$ frequently.

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A shorter alternative solution: use the following computation rule (Proposition 2.4 from this note).

Let $(a_n)_n$ and $(b_n)_n$ be two bounded real sequences such that $b_n$ converges to $b \ge 0$. Then \begin{align} \limsup_n\, a_nb_n &= b \limsup_n\, a_n \text{ and } \\ \liminf_n\, a_nb_n &= b \liminf_n\, a_n. \end{align}

Apply this rule with $a_n = \cos\dfrac{n\pi}{3}$ and $b_n = 1 + \dfrac{1}{2n}$ to get the same conclusion.