Prove that $ne^{-na} \leq C e^{\frac{-na}{2}}$ How to Prove that for any $a > 0$ there exists $C \in R$ such that for all $n \geq 1$
$$ne^{-na} \leq C e^{-na/2}$$
 A: Let $a > 0.$ If $C \in \mathbb{R},$ then $C$ has the property
iff
$$ne^{-na} \leq C e^{-na/2}$$
for all $n \geq 1,$
which holds iff
$$\frac{n}{e^{na/2}} \leq C$$
for all $n \geq 1,$
which holds iff
$$e^{\log n - na/2} \leq C$$
for all $n \geq 1.$
Since there is an $N \geq 1$ such that $\log n < na/2$ for all $n > N,$
so there is a $C \in \mathbb{R},$ 
say 
$$C := \sup \{ e^{\log n - na/2} \mid n \leq N \} + 1,$$
such that $C$ has the property.
So yes, the conjecture is true.
A: Hint: Try taking logs on both sides to arrive at either a condition or contradiction. In this case after taking logs, you get
$$\log C \geq \log n -\frac{na}{2}$$
Now we have $a>0$. Since $n$ grows faster than $\log(n)$, for $n > N(a)$ for some large $N(a)$, the RHS goes negative. Now consider
$$\max_{n=1...N(a)}\left( \log n -\frac{na}{2}\right)$$
A: $\displaystyle  ne^{-na} \leq C e^{-na/2} \Longleftrightarrow \frac{n}{ e^{na/2}} \leq C$
Using the basic inequality $ e^x \geq x+1 $ which is true for every $ x \in \mathbb R$ we obtain:
$\displaystyle \frac{n}{ e^{na/2}} \leq \frac{n}{na/2 +1} = \frac{2n}{na+2} \to \frac{2}{a} $ ,as $ n \to \infty$.
So, you can pick $\displaystyle C=C(a) = \frac{2}{a} $.
