Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Does $$\lim _{n\rightarrow \infty }\left( e^{-na}n^{b}\right) $$ evaluate to $\infty$ when $a > 0, b > 0$.

I tried the expansion of $e^{-na}$ but could not shake of n from numerators.

share|improve this question
    
Think of the expression as $n^b\over e^{na}$ and note all powers are positive. Which wins out, exponentials or powers? –  David Mitra Mar 6 '12 at 19:17
    
it should evaluates to 0, since for a large number exponential is way larger than polynomial. –  quartz Mar 6 '12 at 19:19
add comment

3 Answers

up vote 2 down vote accepted

To try and complete your attempt:

You can use the expansion of $e^n$ to show that for any $c \gt 0$, $e^n \gt Kn^c$ for some constant $K \gt 0$ (dependent on $c$).

Let $[c] = m-1$ ($[x]$ is the integer part of $x$).

Since $e^n = 1+ n + \frac{n^2}{2} + \dots + \frac{n^{m}}{m!} + \dots \gt \frac{n^m}{m!}$

Now $n^m \gt n^c$ and so $e^n \ge \frac{n^c}{m!}$

In your case, we can pick $c = \frac{b+1}{a}$.

So we get

$$e^n \ge K n^{\frac{b+1}{a}}$$

i.e

$$ e^{na} \ge K^a n^{b+1}$$

and

so

$$ \frac{n^b}{e^{na}} \le \frac{1}{K^a n}$$

And so your sequence converges to $0$.

btw, you don't really need the infinite series of $e^x$.

Try proving, by induction on $n$ that: $e^x \ge 1 + x + \frac{x^2}{2!} + \dots + \frac{x^n}{n!}$ for $x \ge 0$.

share|improve this answer
add comment

The answer is no.

Expressing $n$ as $e^{\log n}$ The equation simplifies to $\text{exp}[{b \log n-na}]$

Since $\log n$ does not grow as fast as $n$, it should evaluate to zero as $n\rightarrow \infty$.

share|improve this answer
add comment

To me, this problem just screams for a certain Frenchman's aid:

Let $k$ be an integer greater than or equal to $b$. Then $$ 0\le{e^{-na}n^b}={n^b\over e^{na}}\le {n^k\over e^{na}}. $$ Now evaluate $\displaystyle\lim\limits_{n\rightarrow\infty}{ {n^k\over e^{na}}} =\lim\limits_{x\rightarrow\infty}{ {x^k\over e^{xa}}} $, by applying L'Hôpital's rule $k$-times.

(The first step isn't necessary, but in my opinion makes the write up a bit prettier.)

share|improve this answer
    
Nice, also thanks for using L'Hopital's rule, i needed a reminder of that. –  Hardy Mar 6 '12 at 19:56
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.