Is this application of the squeeze theorem on a sequence correct?

Question

I was requested to decide if $$a_n=\frac{sin^2n}{4^n}$$ was convergent or divergent. I proved it was convergent in the following way.

According to theorem $3$ (this is how it's numerated on my academic text book), if $\lim_{x\to\infty}f(x)=L$ and $f(n)=a_n$, then $\lim_{n\to\infty}a_n=L$.

$i)$ Let $f(x)=\frac{sin^2x}{4^x}$ so that $f(n)=a_n$.

$ii)$ Consider that $$0\leq sin^2x \leq 1 \implies 0\leq f(x) \leq \frac{1}{4^x} \implies \lim_{x\to\infty}0 \leq \lim_{x\to\infty} f(x) \leq \lim_{x\to\infty} \frac{1}{4^x}$$

$$ \implies 0 \leq \lim_{x\to\infty}f(x) \leq 0$$

According to the squeeze theorem, it follows that $\lim_{x\to\infty}f(x)=0$.

$iii)$ According to theorem $3$, then $\lim_{n\to\infty}a_n=0.$ Therefore, the sequence is convergent.

Is this proof formally correct? Can it be improved or simplified? Thank you.

Answer 1

You should rewrite your proof, starting form $ 0 \leqslant f(x) \leqslant \frac{1}{4^x}$, as the following.

Since $ \lim_{x \rightarrow \infty} 0 = 0 $ and $ \lim_{x \rightarrow > \infty} \frac{1}{4^x} = 0$, it follows from the squeeze theorem that $ \lim_{x \rightarrow \infty} f(x) = 0$.

Then the proof continues as you wrote in $iii)$.