Sup and inf of $n \sin(1/n)$

Question

If $n$ is a natural number then, what is the supremum and infimum of $n\sin(1/n)$? is the question I want to solve. I drew $sin(x)/x$ graph and I think that the supremum is $1$ and infimum is $sin(1)$. Is that right? And how can I solve?

Answer 1

You are right. Let $f(x)=x\sin\frac1x$. Note that $f'(x)$ is decreasing.

so the infimum is $f(1)$ and the supremum is $\lim_{n\to \infty}n\sin\frac1n=1$

Answer 2

Let $f(x) = x \sin(\frac{1}{x})$. We are interested in its behaviour for $x \ge 1$.

$f'(x) = \sin(\frac{1}{x}) - \frac{1}{x} \cos(\frac{1}{x})$.

$f''(x) = -\frac{1}{x^3} \sin(\frac{1}{x})$, and this is negative for any $x \ge 1$.

Then, $f'$ is decreasing. But:

$$\lim_{x \to +\infty} f'(x) = 0$$

Then $f'(x) > 0$, for all $x \ge 1$. So $f$ is increasing on $[1, +\infty[$.

Therefore the sequence $n\sin(\frac{1}{n})$ is increasing for all $n \ge 1$. Therefore:

$$\inf_{n \in \mathbb N} \{n\sin(\frac{1}{n}) \}= 1 \times \sin(\frac11) = \sin(1)$$

$$\sup_{n \in \mathbb N} \{n \sin(\frac{1}{n})\} = \lim_{n \to +\infty} n \sin(\frac{1}{n}) = \lim_{n \to +\infty} \frac{\sin(\frac{1}{n})}{\frac{1}{n}} = 1$$