$^n\sqrt{2^n-n^2} = ?$

Question

I think that $\sqrt[n]{2^n-n^2} = 2$

I've tried to use the squeeze theorem for sequences: i.e $\sqrt[n]{2^n-n^2} \le \sqrt[n]{2^n} = 2$, but I couldn't find a lower bound. Any idea?

Thanks.

Answer 1

Note that

$$\sqrt[n]{2^n-n^2}=\sqrt[n] {2^n}\cdot \sqrt[n]{1-\frac{n^2}{2^n}}=2\cdot \sqrt[n]{1-\frac{n^2}{2^n}}\to 2\cdot 1 =2$$

indeed

$$\sqrt[n]{1-\frac{n^2}{2^n}}=e^{\frac{\log \left(1-\frac{n^2}{2^n}\right)}{n}}\to e^0=1$$

Answer 2

HINT: $\frac12\cdot2^{n}=2^{n-1}>n^2$ for large $n$.