Why $\lim_{n \to \infty} \frac{2^n}{\sqrt{n}^\sqrt{n}} = \infty$?

Question

Could you please explain to me why this is true:

$$\lim_{n \to \infty} \frac{2^n}{\sqrt{n}^\sqrt{n}} = \infty$$

I have tried L'Hôpital's rule, but it doesn't add any clarification.

Answer 1

The series $$\sum {\sqrt{n}^{\sqrt{n}}\over 2^n}$$ is convergent due to the $n$-th root test. Indeed we get $${{\sqrt{n}^{1\over\sqrt{n}}\over 2}}\to{1\over 2}$$ Hence the general term tends to $0,$ and its reciprocal to $+\infty.$

On the way we have used a well known fact that $\lim_{x\to\infty}x^{1/x}=1$ which follows easily from $\lim_{n\to\infty}(n+1)^{1/n}=1$ basing on $(1+\delta)^n\ge {n(n-1)\over 2}\delta^2,$ for $\delta>0.$

Answer 2

Hint.

$\bullet~ \frac{2^{n}}{\sqrt n ^{\sqrt n}} = \left( \frac{2^{\sqrt{n}}}{\sqrt n}\right)^{\sqrt{n}}$

$\bullet$ $2^x > 2x$ holds for $x\geq 2$.

Answer 3

$\begin{array}\\ \dfrac{2^n}{\sqrt{n}^\sqrt{n}} &=\dfrac{2^n}{(n^{1/2})^\sqrt{n}}\\ &=\dfrac{2^n}{n^{\sqrt{n}/2}}\\ &=\left(\dfrac{2}{n^{1/\sqrt{n}}}\right)^n\\ &=\left(\dfrac{2}{\sqrt{n}^{2/\sqrt{n}}}\right)^n\\ &=\left(\dfrac{2}{\left(\sqrt{n}^{1/\sqrt{n}}\right)^2}\right)^n\\ \end{array} $

and since

$(1+n^{-1/2})^n \ge 1+n^{1/2} \gt n^{1/2} $ (by Bernoulli) so, raising to the $2/n$ power, $n^{1/n} \lt (1+n^{-1/2})^2 \lt 1+3n^{-1/2} $ so, putting $n^{1/2}$ for $n$,

$\sqrt{n}^{1/\sqrt{n}} \lt 1+3n^{-1/4} \lt 1+\frac14$ for large enough $n$ so

$\dfrac{2^n}{\sqrt{n}^\sqrt{n}} \gt \left(\dfrac{2}{(5/4)^2}\right)^n =\left(\dfrac{32}{25}\right)^n \to \infty $

for large enough $n$.