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.
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 $$ \lim_{N\to \infty}\left(1+\frac{x}{N}\right)^{N}=e^x. $$
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 $$ \lim_{n\to \infty} \left(1+\frac{1}{3n+1}\right)^{3n-1} = \lim_{n\to \infty} \left(1+\frac{1}{3n+1}\right)^{3n+1} \lim_{n\to \infty} \left(1+\frac{1}{3n+1}\right)^{-3}=e\times 1 =e. $$
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!
