How to prove that the sequence $a_{n}=2^{n}\times \sin(\frac{1}{n})$ diverges as $n$ approaches $\infty$?

Question

Trying to choose $2$ sub sequences that converge to a different limit to show that an diverges but i'm not succeeding,any tips?

Answer 1

Hint: For $\frac 1n< \frac{\pi}{3}$ you have $\sin\left(\frac 1n\right)\geq \frac{1}{2n}$ by the mean value theorem.

Answer 2

For $0 < x < \dfrac{\pi}{3} \implies \sin x > \dfrac{x}{2}$. This is true because $f(x) = \sin x - \dfrac{x}{2}$ has $f'(x) = \cos x - \dfrac{1}{2} > 0\implies f(x) > f(0) = 0\implies \sin x > \dfrac{x}{2}$. Put $x = \dfrac{1}{n}$, the answer follows since $\dfrac{2^n}{n} \to \infty$ as $n \to \infty$.