# Show that for all $\lambda \geq 1~$ $\frac{\lambda^n}{e^\lambda} < \frac{C}{\lambda^2}$

Show that for any $n \in \mathbb N$ there exists $C_n > 0$ such that for all $\lambda \geq 1$

$$\frac{\lambda^n}{e^\lambda} < \frac{C_n}{\lambda^2}$$

I can see that both sides of the inequality have a limit of $0$ as $\lambda \rightarrow \infty$ since, on the LHS, repeated application of L'Hôpital's rule will render the $\lambda^n$ term as a constant eventually, while the $e^{\lambda}$ term will remain, and the RHS is obvious.

I can also see that the denominator of the LHS will become large faster than the RHS denominator, but I can't seem to think of anything that will show that the inequality is true for all the smaller intermediate values.

-
Hint: take reciprocals and use the power series expansion of $\exp$ – user20266 Jun 9 '12 at 6:59
I have tried to rearrange your quantifiers so that the question makes sense. Let me know if I misunderstood you! – TonyK Jun 9 '12 at 9:37

HINT: The inequality $$\frac{\lambda^n}{e^\lambda} < \frac{C}{\lambda^2}$$ is equivalent to the inequality $\lambda^{n+2}e^{-\lambda}<C$. Consider the function $f(x)=x^{n+2}e^{-x}$ on the positive real line; does it have a maximum somewhere?

-
Beat me to it :-) (+1) – robjohn Jun 9 '12 at 7:07
When $(n + 2) - x = 0$? I'm not sure how to deal with the $n$ term exactly. I'm guessing the numerator would be smallest when $n = 0$, and therefore at the function's max, right? assuming $0 \in \mathbb{N}$ Can I just take $n$ to = $0$ and work with that? – stariz77 Jun 9 '12 at 7:19
@stariz77: vary $\lambda$; leave $n$ fixed. – robjohn Jun 9 '12 at 7:28
@stariz77: Yes, when $x=n+2$. Thus, for any given $n$ the maximum is $(n+2)^{n+2}e^{-(n+2)}$. You can prove that this tends to $\infty$ as $n\to\infty$, so you can’t hope to find a $C$ that works for all $n$ simultaneously, if that’s what you’re trying to do. – Brian M. Scott Jun 9 '12 at 7:29

The function $\lambda \mapsto \lambda^{n+2}$ is strictly increasing for positive $\lambda$ and also $e^{\lambda} > \lambda$. Combining this you get

$$e^{\lambda} = \left( e^{\frac{\lambda}{n+2}} \right)^{n+2} > \left( \frac{\lambda}{n + 2} \right)^{n+2}$$

for all positive $\lambda$ and therefore

$$\frac{\lambda^n}{e^\lambda} < \frac{\lambda^n (n+2)^{n+2}}{\lambda^{n+2}} = \frac{(n+2)^{n+2}}{\lambda^2}.$$

-

Considering the power series of exp seems to give the simplest solution. For $\lambda>0$ we have $$e^\lambda = \sum_{k=0}^{\infty} \frac{\lambda^k}{k!}$$ $$e^\lambda > \frac{\lambda^{n+2}}{(n+2)!}$$ $$\frac{(n+2)!}{\lambda^2} > \frac{\lambda^{n}}{e^\lambda}$$ so that $C=(n+2)!$. As Brian M. Scott remarks in a comment above, $C$ necessarily depends on $n$ as $\lambda^n/e^\lambda\rightarrow\infty$ as $n\rightarrow \infty$ for $\lambda > 1$.

-