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$$\lim_{n\rightarrow\infty}\ a_n$$

$$a_n= \frac{3n}{5+3^{n+1}}$$

I think I should use the squeeze theorem to solve this problem, but I have trouble finding the upper bound. Could the lower bound be $$b_n= \frac{3n}{n(5+3^{n+1})}$$?

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    $\begingroup$ What is $ \lim_{x\to \infty}\dfrac{x}{e^x}$? $\endgroup$ Feb 25 '18 at 12:41
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We have

$$0 \leq \frac{3n}{5+3^{n+1}} \leq \frac{3n}{3^{n+1}} = \frac{n}{3^n}$$

As $\displaystyle \lim_{n \to \infty} 0 \quad$ and $\quad \displaystyle \underbrace{\lim_{n \to \infty} \frac{n}{3^n}=0}_{\text{Why?}}$

By squeeze theorem, $\displaystyle \lim_{n \to \infty}a_n=0$

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Prove by induction that $3^{n+1}>n^2$.

Thus, $$0<a_n<\frac{3n}{5+3^{n+1}}<\frac{3n}{5+n^2}=\frac{\frac{3}{n}}{\frac{5}{n^2}+1}\rightarrow0.$$ Now, use sandwich.

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Or use L'Hôpital's rule:

$$\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac 1 {3^n\log 3}=0$$

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By ratio test

$$\frac{a_{n+1}}{a_n}=\frac{3(n+1)}{5+3^{n+2}}\cdot\frac{5+3^{n+1}}{3n}=\frac{n+1}{n}\frac{}{}\frac{5+3^{n+1}}{5+3\cdot3^{n+1}}\to \frac13$$

thus

$$a_n\to 0$$

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