Evaluation of $\lim _{n\rightarrow \infty }\left( e^{-na}n^{b}\right) $ when $a > 0, b > 0$ 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. 
 A: 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 $n\rightarrow \infty$.
A: 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 $x \ge 0$.
A: 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.)
