If $ \lim_{n\to \infty} \sqrt[n]{a_n} = e$ then the sequence $(a_n \cdot 3 ^{-n})$ converges to $0$ My Thoughts:
I think the statement is correct. Please fix me if I'm wrong.
I know that $2< e< 3$, by powering by $n$ I get $2^n< e^n< 3^n$
Now, I fulfill the IF condition; The limit of it is $e$.
and limit of  $(a_n  \cdot 3 ^{-n})$ converges to $0$ because $ ({e \over 3})^n$ and I know that if the expressions is smaller than 1 when $n$ tends to infinity it converges to $0$
 A: If $\lim_{n\to \infty} \sqrt[n]{a_n} = e$, then there is an $n_0$ such that $\sqrt[n]{a_n}<\frac{e+3}{2}$ for $n\ge n_0$. Thus:
$$a_n3^{-n}\le\left(\frac{e+3}{6} \right)^n \to 0$$
(there are, in principle, details to cover re: whether $a_n$ is positive or not; these are not difficult to incorporate)
A: Consider the sum $\sum a_n 3^{-n}$. I claim that this converges, and I use the root test to prove this. 
$$\lim_{n \to \infty}  \ ({|a_n| 3^{-n}})^{\frac 1n} = \lim_{n \to \infty} \frac{1}{3} |a_n|^{\frac 1n}$$
It's given that $\lim_{n \to \infty} a_n^{\frac 1n} = e$. This implies that $a_n$ is nonnegative for sufficiently large $n$ (otherwise the expression would be undefined if $n$ is even, and negative if $n$ is odd, and therefore not "close" to $e$). 
In other words, for sufficiently large $n$ ,$|a_n| = a_n.$
Therefore, 
$$\lim_{n \to \infty} \frac{1}{3} |a_n|^{\frac 1n} = \lim_{n \to \infty} \frac{1}{3} a_n^{\frac 1n} = \frac{e}{3} < 1$$
Since $\sum a_n 3^{-n} < \infty$, $a_n3^{-n} \to 0$. 
A: The crucial idea here is that the limit $\lim_{n \to \infty}a_{n}^{1/n} = e$ is less than $3$ and hence $a_{n}\cdot 3^{-n} \to 0$. More generally $$a_{n}^{1/n} \to a > 0, a < b \Rightarrow a_{n}/b^{n} \to 0$$ The idea is to chose a number $c$ between $a, b$ i.e. $a < c < b$. Since $a_{n}^{1/n} \to a$ it follows that $$a_{n}^{1/n} < c$$ for all $n \geq m$ and then $$a_{n} < c^{n}$$ and therefore $$0 \leq \frac{a_{n}}{b^{n}} < \frac{c^{n}}{b^{n}} = \left(\frac{c}{b}\right)^{n}$$ for all $n \geq m$. Since $c < b$ it follows that $0 < c/b < 1$ and hence $(c/b)^{n} \to 0$ and therefore applying Squeeze theorem we get $$\lim_{n \to \infty}a_{n}b^{-n} = 0$$
