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

Question

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.

2. 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.

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

Answer 1

$\frac {\sin x} x \to 1$ 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 $n \to \infty$.

Answer 2

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:

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