# prove the following series doesn't converge : $a_n = 2^n \cdot \sin(\frac 1n)$

I'm reviewing my Calculus material, and I'm in the sub-sequences chapter. In my book there is this exercise where I need to prove, using theorems and lemmas related to the topic, that the following sequence does not converge: $$a_n = 2^n \cdot \sin(\frac 1n)$$.

My attempts : 1. I tried finding 2 sub-sequences that converge to different limits. No success here.

1. I tried proving $$\lim_{ n\to \infty} a_n = \infty$$ using :
"if for every n $$\in \Bbb N : a_n>b_n$$ and $$\lim_{n \to \infty} b_n = \infty$$ , then $$\lim_{n \to\infty}a_n=\infty$$ " But I couldn't find a good $$b_n$$. I also tried using here the GM-HM inequality but it didn't yield a good result:

$$2^n\cdot \sin(\frac 1n) \ge \frac{4}{\frac{1}{2^n}+\frac{1}{\sin(\frac 1n)}} \to_{n \to \infty} = 0$$

Another note : I also tried proving the sequence is strictly Monotone but I couldn't figure out how to do so ? Because if I prove this and show it is unbounded from above, then I know $$a_n$$ diverges.

$$\frac{a_{n+1}}{a_n} = 2\cdot \frac{\sin(\frac{1}{n+1})}{\sin(\frac 1n)}$$

$$\frac {\sin x} x \to 1$$ as $$x \to 0$$. Hence $$\frac {a_n} {b_n} \to 1$$ where $$b_n=\frac {2^{n}} n$$. You can see that $$b_n \to \infty$$ and conclude that $$a_n \to \infty$$?

To, show that $$\frac {\sin x} x \to 1$$ as $$x \to 0$$ without L'Hopital's Rule you can use the Taylor series expansion of $$\sin\, x$$.

Let $$M$$ be any positive number. Let $$n >2M-1$$. Note that $$2^{n} >n+\frac {n(n-1)} 2$$ by Binomial Theorem. This gives $$2^{n}> Mn$$ proving that $$\frac {2^{n}} n \to \infty$$ as $$n \to \infty$$.

• Thank you for the quick answer ! I failed to understand how $b_n \to \infty \Rightarrow a_n \to \infty$ . Could you please elaborate ? – Shahar Palensya Aug 1 at 6:37
• $\frac {b_n} {a_n} >\frac 1 2$ for $n$ sufficiently large. Hence $a_n >\frac 1 2 b_n$ from which it follows that $a_n \to \infty$. – Kabo Murphy Aug 1 at 6:40
• Trying to actually show $\sin x/x\to1$ is fraught with perils and circular arguments. Whether the Taylor expansion is a valid argument depends entirely on how you find said Taylor expansion. If it's your definition of $\sin x$, fine. If not, then you usually find it by differentiating $\sin x$, which you can't do without already knowing said limit. – Arthur Aug 1 at 6:40
• @Arthur What properties of sine function we can use in the answer is something that that OP has to decide. Most of the time we do know much about the level of understanding of the OP, we have to gamble a bit. You can now see that OP is happy with my answer in this case. – Kabo Murphy Aug 1 at 7:16
• @KaviRamaMurthy I largely agree, but on one specific point, you don't have to gamble. You can use the comments to ask for clarification from the asker about how they understand $\sin$. – Theo Bendit Aug 1 at 8:00

Maybe you know already the following inequality

• $$\sin x \geq \frac{2}{\pi}x$$ for $$0\leq x \leq \frac{\pi}{2}$$ Then you get immediately:

$$a_n = 2^n \cdot \sin(\frac 1n)\geq \frac{2}{\pi}\frac{(1+1)^n}{n}>\frac{2}{\pi}\frac{\frac{n(n-1)}{2}}{n}= \frac{n-1}{\pi}\stackrel{n \to \infty}{\longrightarrow} +\infty$$