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How to Prove that for any $a > 0$ there exists $C \in R$ such that for all $n \geq 1$

$$ne^{-na} \leq C e^{-na/2}$$

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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.

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  • $\begingroup$ Chou, in your final line, why have you added $+1$. Is it not sufficient to set $C$ as the least upper bound? $\endgroup$ – elbarto May 28 '15 at 14:10
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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 $$\max_{n=1...N(a)}\left( \log n -\frac{na}{2}\right)$$

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$\displaystyle ne^{-na} \leq C e^{-na/2} \Longleftrightarrow \frac{n}{ e^{na/2}} \leq C$

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} $.

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