Finding Limit of n-th root sequence Disclaimer: I'm a starting-level-student.
How do I prove or deny this?
 $$\lim_{n\to \infty} \sqrt[n]{2^n - n^2} = 2$$
I have a feeling that this expression doesn't getting closer to 2 but I can't find how to prove this.
I've tried 2 ways:
First - 
$$|\sqrt[n]{2^n - n^2} - 2| < \epsilon$$ 
from the definition of limits, trying to evaluate this expression and deny it but I'm not getting anywhere.
So can I get some help on how to solve this ?
 A: HINT: write your Limit in the form $$2\sqrt[n]{1-\frac{n^2}{2^n}}$$
A: You may exploit the following inequality: if $a>b>0$, for any $n\geq 1$ we have
$$ n(a-b)b^{n-1}\leq a^n-b^n \leq n(a-b)a^{n-1}\tag{1}$$
Proof: it simply follows from $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\ldots+a b^{n-2}+b^{n-1})$. There are $n$ terms of the form $a^j b^k$ and each of them is $\leq a^{n-1}$ and $\geq b^{n-1}$due to $b<a$.
Now we may consider $a=2$ and $b=\left(2^n-n^2\right)^{1/n}$. We get
$$ 2-(2^n-n^2)^{1/n}\leq \frac{n^2}{n \left(2^n-n^2\right)^{\frac{n-1}{n}}}\leq \frac{n}{2^{n/2}}\tag{2} $$
where the last inequality holds for any $n$ large enough. Now $(2)$ clearly proves that
$$ \lim_{n\to +\infty}\left(2^n-n^2\right)^{1/n}=2 \tag{3} $$
as wanted.

$(1)$ might look as an overkill, but it is a useful inequality for proving, for instance,
$$ \lim_{n\to +\infty}\int_{0}^{n}x^n\left(1-\frac{x}{n}\right)^n\,dx = \int_{0}^{+\infty}x^n e^{-x}\,dx = n! $$
leading to Euler's product formula for the $\Gamma$ function.
