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What are supremum and infimum of $\{nsin\frac{1}{n}\}_{n=1}^\infty$? And is maximum and minimum assumed?

I would appreciate help with the task above, I know how to find the supremum and infimum, but I do not manage to determine if the function is increasing or decreasing to see if a maximum and/or a minimum value exists.

I let $f(x)=xsin\frac{1}{x}$.

I have tried to determine if the function is increasing or decreasing with $f^\prime(x)=\sin\frac{1}{x}-\frac{1}{x}\cos\frac{1}{x}$ but it is not possible to solve this when equaled to zero so I cannot check the intervals around these x-values.

I have also tried reasoning that $n$ increases faster than $sin\frac{1}{n}$ decreases and therefore f(x) increases, but I cannot find a way to prove this.

Thanks in advance!

kabin
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Note that $n \sin(1/n) \to 1$, as we can write it as $\frac{\sin(1/n)}{1/n}$ as $1/n \to 0$. Further, since $\sin(x) < x$ for all $x > 0$, we see that $\sin(1/n) < 1/n$, hence $n\sin(1/n) < 1$. This tells me that the supremum is $1$, but a maximum is not achieved.

The infimum is going to be a minimum, achieved at some sequence point. Your derivative calculations reveal that $$f'(x) = \sin\left(\frac{1}{x}\right) - \frac{1}{x}\cos\left(\frac{1}{x}\right) = \cos\left(\frac{1}{x}\right)\left(\tan\left(\frac{1}{x}\right) - \frac{1}{x}\right).$$ As $\tan(x) > x$ for $0 < x < \pi/2$, we have $f'(x) > 0$ for $0 < x < 2/\pi$. So, for $n \ge 2$, this means the sequence is increasing. The minimum must occur either at $n = 1$ or $n = 2$. Comparing these values, the minimum is $\sin(1)$, i.e. at $n = 1$.

Theo Bendit
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  • Thank you very much! But I only have one question, would it also be correct to assume after we have established that sup = 1, and that no maximum exists, that the function is increasing? And only point out that inf = min is obtained when n = 1? – kabin Sep 13 '21 at 00:34
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    Yes, that's right. As I mentioned in my answer, the supremum is not achieved (i.e. the maximum doesn't exist) because $n\sin(1/n) < 1$, but the limit is $1$. The only point where the infimum could be achieved is $n = 1$ or $n = 2$, but since $\sin(1) < 2\sin(1/2)$, the infimum is achieved only at $n = 1$. – Theo Bendit Sep 13 '21 at 01:43
  • Thank you very much for your answer, it was very easy to understand and informative! – kabin Sep 13 '21 at 21:53