Proving that this function is negligible Let $f(n) =\frac{1}{2^{\sqrt n}}$, where $n \in \mathbb{N}$.
I want to prove that $\forall a \in N, a \ge1\; \exists k: f(n) \le n^{-a}, \forall n \ge k$
I attempted to solve the inequality $\frac{1}{2^{\sqrt n}} \le \frac{1}{n^a}$, but I got nowhere. Can someone show me how to solve it?
EDIT. Clarification: I want to find the minimum value of k (in function of a). Proving that the previous inequality is true for certain values of n is not enough.
 A: Since $e^{x/2}<2^x$ for $x>0$, by Taylor expansion of $e^{x/2}$ about $x=0$, we obtain the inequality $$\dfrac{n^{a+1}}{2^{2a+2}(2a+2)!}<\sum_{k=0}^{2a+2}\dfrac{(\sqrt{n}/2)^k}{k!}<e^{\sqrt{n}/2}<2^{\sqrt{n}}$$
So, it suffices to show that there exists some $k$ such that $\forall n\geq k$ $2^{2a+2}(2a+2)!\leq n$, but this obviously holds, since $n$ diverges to positive $\infty$, whereas the former is a constant. 
A: Note that
$$\frac1{2^{\sqrt{n}}} \leqslant n^{-a} \iff  (\ln 2)\sqrt{n} \geqslant a\ln n.$$
Consider
$$f(x) = \frac{(\ln 2)\sqrt{x}}{a \ln x}$$
Then, using L'Hospital's rule,
$$ \lim_{x \rightarrow \infty}f(x) = \lim_{x \rightarrow \infty}\frac{\ln 2}{2a} \sqrt{x}=\infty,$$
and $f(x) \geqslant 1$ for all sufficently large $x$.
Whence, for sufficiently large $n$
$$ (\ln 2)\sqrt{n} \geqslant a\ln n$$
To find the minimum $k$ such that the inequality holds, solve
$$g(x) = (\ln 2)\sqrt{x} - a\ln x = 0.$$
Note that $g$ is increasing for $x \geq (2a/\ln 2)^2.$
The solution involves the Lambert W function. 
