# $\lim_{n \rightarrow \infty} \frac{(-1)^{n}\sqrt{n}\sin(n^{n})}{n+1}$. Am I correct?

I have to find this limit:

\begin{align} \lim_{n \rightarrow \infty} \frac{(-1)^{n}\sqrt{n}\sin(n^{n})}{n+1} \end{align}

My attempt:

Since we know that $$-1\leq \sin (x) \leq 1$$ for all $$x \in \mathbb{R}$$ we have:

\begin{align} \lim_{n \rightarrow \infty} \frac{(-1)^{n}\cdot\sqrt{n}\cdot(-1)}{n+1}\underbrace{\leq}_{\text{Is this fine?}}&\lim_{n \rightarrow \infty} \frac{(-1)^{n}\cdot\sqrt{n}\cdot\sin{(n^{n})}}{n+1} \leq \lim_{n \rightarrow \infty} \frac{(1)^{n}\cdot\sqrt{n}\cdot(1)}{n+1}\\ \\ \lim_{n \rightarrow \infty} \frac{(-1)^{n+1}\cdot\sqrt{n}}{n+1}\leq&\lim_{n \rightarrow \infty} \frac{(-1)^{n}\cdot\sqrt{n}\cdot\sin{(n^{n})}}{n+1} \leq \lim_{n \rightarrow \infty} \frac{\sqrt{n}}{n+1} \end{align}

My doubt in the inequeality is because of the $$(-1)$$ terms. If everything is correct, then we have

\begin{align} \lim_{n \rightarrow \infty} \frac{(-1)^{n+1}\cdot\sqrt{\frac{1}{n}}}{1+\frac{1}{n}}\leq&\lim_{n \rightarrow \infty} \frac{(-1)^{n}\cdot\sqrt{n}\cdot\sin{(n^{n})}}{n+1} \leq \lim_{n \rightarrow \infty} \frac{\sqrt{\frac{1}{n}}}{1+\frac{1}{n}}\\ \\ \Rightarrow \ \ \ 0 \leq &\lim_{n \rightarrow \infty} \frac{(-1)^{n}\cdot\sqrt{n}\cdot\sin{(n^{n})}}{n+1} \leq 0 \\ \\ \therefore& \lim_{n \rightarrow \infty} \frac{(-1)^{n}\cdot\sqrt{n}\cdot\sin{(n^{n})}}{n+1}=0 \end{align}

Am I correct? If I am not, how can I solve it? There are other ways to find this limit? I really appreciate your help!

• To make the argument a bit easier, you can use the fact that $\lim_{n\to\infty}|a_n|=0$ implies $\lim_{n\to\infty} a_n=0$. This will eliminate the $(-1)^n$ term entirely. Commented Jan 5, 2021 at 20:22

The first inequality is not true in general: if $$a\leqslant b$$ and $$c<0$$, then $$ac\color{red}{\geqslant}bc$$.

But note that$$\left|\frac{(-1)^{n}\sqrt{n}\sin(n^{n})}{n+1}\right|\leqslant\frac{\sqrt n}{n+1}\to0,$$and therefore your limit is $$0$$ indeed.

Fact 1. If $$a_n\leq b_n$$ for large $$n$$ then $$\displaystyle\lim_{n\to\infty}a_n\leq\lim_{n\to\infty}b_n$$, provided the limits exist.

Fact 2. If $$\displaystyle\lim_{n\to\infty}|a_n|=0$$ then $$\displaystyle\lim_{n\to\infty}a_n=0$$.

From fact 1 we have $$0\leq\lim_{n\to\infty}\left|\frac{(-1)^n\sqrt{n}\sin(n^n)}{n+1}\right| \leq\lim_{n\to\infty}\frac{\sqrt{n}}{n+1} =0$$ and so we can apply fact 2 to obtain $$\lim_{n\to\infty}\frac{(-1)^n\sqrt{n}\sin(n^n)}{n+1} =0.$$

\begin{align} \lim_{n \rightarrow \infty} \frac{(-1)^{n}\sqrt{n}\sin(n^{n})}{n+1} \end{align}

This is one of those problems where you are supposed to take a step back and make things easy on yourself.

Let

$$f(n) = (-1)^{n}\sin(n^{n})$$

and let

$$g(n) = \frac{\sqrt{n}}{n+1} < \frac{1}{\sqrt{n}}.$$

You want

$$\lim_{n\to \infty} [f(n) \times g(n)].$$

As $$n \to \infty, f(n)$$ is a bounded function.
That is $$~-1 \leq f(n) \leq +1~$$ for all $$n$$.

Also, clearly, as $$n \to \infty, g(n) \to 0.$$

Therefore, as $$n \to \infty, \{f(n) \times g(n)\} \to 0.$$

Edit
For all $$n, |f(n) \times g(n)| \leq |g(n)|$$ which goes go zero.

Therefore, since $$|f(n) \times g(n)| \to 0,~ \{f(n) \times g(n)\} \to 0.$$

• My English it is very poor (I have written limited) :-( Commented Jan 5, 2021 at 20:37
• @Sebastiano +1 for your answer, which I regard as valid. Commented Jan 5, 2021 at 20:40
• Ahahah :-) there is not problem for the vote for me :-). However thank you. My version is the one I do in high school. Unfortunately, when I have to do with the upper-bound, the students don't understand what it means because they haven't understood that inequalities are needed. Commented Jan 5, 2021 at 20:44

If $$(a_n)$$ is a bounded sequence and $$(b_n)$$ is an infinitesimal sequence, then the product sequence $$(a_n \cdot b_n)$$ is also infinitesimal.

$$\lim_{n \rightarrow \infty} \frac{(-1)^{n}\sqrt{n}\sin(n^{n})}{n+1}$$

Being $$\forall n\in \mathbb{N}$$ then $$-1 \leq (-1)^{n} \le 1$$ and $$-1\leq \sin (n^n)\leq 1$$ two bounded sequences then $$\lim_{n \rightarrow \infty} \frac{(-1)^{n}}{n+1}=\lim_{n \rightarrow \infty} \frac{\sin (n^n)}{n+1}=0$$

$$\lim_{n \rightarrow \infty} \underbrace{[(-1)^{n}\sin(n^{n})]}_{\text{bounded sequences}}\,\frac{\sqrt{n}}{n+1}=\lim_{n \rightarrow \infty} [(-1)^{n}\sin(n^{n})]\cdot \underbrace{\sqrt{\frac{n}{(n+1)^2}}}_{\text{infinitesimal sequence}}=0$$

You're either right or very nearly right: in either case I would be more comfortable putting some absolute value signs on the function to get rid of the $$(-1)^n$$ terms.

Here's why: let's say we have some $$f(x)$$ such that $$\lim_{x \to a} |f(x)| = 0$$. Notice by the linearity of limits that $$\lim_{x \to a} -|f(x)| = -\lim_{x \to a} |f(x)| = 0$$.

So, because for any $$f(x)$$ we have $$-|f(x)| \leq f(x) \leq |f(x)|$$, by the squeeze theorem we have that $$\lim_{x \to a} f(x) = 0$$.

So, the $$(-1)^n$$ terms aren't an issue, but I would state this explicitly in your work.