Problem in solving a question related to Sandwich theorem. The question is :

Show that by Sandwich theorem the sequence  $\left\{\left(1 + \frac{1}{3n+1}\right)^{3n} \right\}_n$ converges to $e$.

Now,what I have done is that $\left(1 + \frac{1}{3n+1}\right)^{3n} < \left(1 + \frac{1}{3n+1}\right)^{3n+1}$.But I fail to construct another part of the inequality.So,Please help me.Thank you in advance.
 A: One may observe that, as $n \to \infty$,
$$
\left(1+\frac1{3n+2}\right)^{3n}\le \left(1+\frac1{3n+1}\right)^{3n}\le\left(1+\frac1{3n+1}\right)^{3n+1}
$$ or $$
\frac1{\left(1+\frac1{3n+2}\right)^2}\left(1+\frac1{3n+2}\right)^{3n+2}\le \left(1+\frac1{3n+1}\right)^{3n}\le\left(1+\frac1{3n+1}\right)^{3n+1}
$$ then conclude with the sandwich theorem, using $\displaystyle \lim_{n \to \infty}\frac1{\left(1+\frac1{3n+2}\right)^2}=1$ and using
$$
\lim_{N\to \infty}\left(1+\frac{x}{N}\right)^{N}=e^x.
$$
A: Try
$$
\left(1+\frac{1}{3n+1}\right)^{3n-1} \le \left(1+\frac{1}{3n+1}\right)^{3n}\le \left(1+\frac{1}{3n+1}\right)^{3n+1}
$$
such that 
$$
\lim_{n\to \infty} \left(1+\frac{1}{3n+1}\right)^{3n+1} = e,
$$
and
$$
\lim_{n\to \infty} \left(1+\frac{1}{3n+1}\right)^{3n-1} = \lim_{n\to \infty} \left(1+\frac{1}{3n+1}\right)^{3n+1} \lim_{n\to \infty} \left(1+\frac{1}{3n+1}\right)^{-3}=e\times 1 =e.
$$
A: First, we can write the term of interest as
$$\left(1+\frac{1}{3n+1}\right)^{3n}=\left(\left(1+\frac{1}{3n+1}\right)^{3n+1}\right)^{3n/(3n+1)}$$
Then noting that $1-\frac{1}{n}<\frac{3n}{3n+1}<1$, we can write
$$\left(\left(1+\frac{1}{3n+1}\right)^{3n+1}\right)^{1-\frac{1}{n}}\le \left(1+\frac{1}{3n+1}\right)^{3n}\le \left(1+\frac{1}{3n+1}\right)^{3n+1} \tag 1$$
Inasmuch as the left-hand and right-hand sides of $(1)$ approach $e$ as $n\to \infty$, the squeeze theorem guarantees that 
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\left(1+\frac{1}{3n+1}\right)^{3n}=e}$$
as was to be shown!
