# Show that there exists a constant $C > 0$, such that for every $n \in \mathbb{N}$ we have $\frac{n^2}{e^n} \leq \frac{C}{n^2}$

I've been stuck on this problem for quite a while, not sure how to proceed.

Show that there exists a constant $C > 0$, such that for every $n \in \mathbb{N}$:

$$\frac{n^2}{e^n} \leq \frac{C}{n^2}$$

I've tried representing the leftside as a powerseries using the exponential function, so that we have:

$$\frac{n^2}{e^n} = n^2 \sum^\infty _{k=0} \frac{(-n)^k}{k!} \leq \frac{C}{n^2}$$

Supposedly $C = 4!$ is a solution, but I'm not quite seeing how to get there.

Any help is greatly appreciated!

Hint: Find the maximum of the function $$f(x)=x^4e^{-x}$$

You should get that any $$C \geq \frac{4^4}{e^4}$$ is a solution.

Since

$$e^n = \sum_{k=0}^\infty \frac{n^k}{k!} > \frac{n^4}{4!},$$

the inequality with $C=4!$ falls right out.

We can also show that $n^4 \leqslant Ce^n$ for $C = 4!$. Consider the taylor series of $e^x$, then \begin{align} 4!(1+n+1/2n^2+1/3!n^3+1/4!n^4 + \ldots) \geqslant n^4 \end{align} for all $n \in \mathbb{N}$.

Hint: multiply each side by $n^2$, noting that it is always positive, and notice that the denominator $e^n$ grows exponentially (do this by comparing their derivatives) and that there exists $a \in \mathbb{R}$ such that $\forall n > a$, $e^n > n^4$. Therefore after a certain $n$, the expression is monotonically decreasing and so given that the terms for $n<a$ are finite, there exists an upper bound. You are only asked to prove there exists a $C$, not to find it. Given this framework, it should be easy to construct a proof.