limit of $n^2(\frac{1+n}{3n})^n$ I want to determine the limit of this succession, for this I suppose that  $$n^2(\frac{n}{3n})^n≤n^2(\frac{1+n}{3n})^n≤n^2(\frac{2n}{3n})^n$$
And as $n^2(\frac{2n}{3n})^n$ and $n^2(\frac{n}{3n})^n$ converge to $0$, then by the squeeze theorem $n^2(\frac{1+n}{3n})^n$ also does
Is this true?
Any help is appreciated
 A: This first comment no longer applies, the post was edited
Your second inequality is not correct. Take for example $n=2$. Then
$$n^2\left(\frac{1+n}{3n}\right)^n=1, \quad \left(\frac{2n}{3n}\right)^n=\frac{4}{9},$$
and clearly $1\geq \frac{4}{9}$.
This is still relevant
What you could do, however, is notice that you can rewrite it as
$$n^2\left(\frac{1+n}{3n}\right)^n=n^2 \left(\frac{1}{3n}+\frac{1}{3}\right)^n=\frac{n^2}{3^n} \left(1+\frac{1}{n}\right)^n.$$
Now notice that
$$\lim_{n\to\infty}\frac{n^2}{3^n}=0$$
(I will assume you already know this), and
$$\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n=e.$$
Thus if you take the limit, you can split it up into a product of limits, and what you end up with is
$$\lim_{n\to\infty}n^2\left(\frac{1+n}{3n}\right)^n=0\cdot e=0.$$
A: You have $n^2\left(\frac{1+n}{3n}\right)^n=\frac{n^2}{3^n}\left(1+\frac{1}{n}\right)^n$.
It is well-known that $\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$ converges to $e$ and it can be shown that $\lim_{n\to\infty}\frac{n^2}{3^n}=0$ (for example, by using L'hospital's Rule twice).
Therefore, $\lim_{n\to\infty} n^2\left(\frac{1+n}{3n}\right)^n=0$.
A: Let $n>1.$
$0<n^2 \left (\dfrac{1+n}{3n}\right)^n <n^2\left(\dfrac{2n}{3n}\right) ^n=$
$n^2\left(\dfrac{2}{3}\right)^n = \dfrac{n^2}{a^n}$, where $a:=3/2>1.$
Take the limit.
