Show that if $k \in \mathbb{N}$ is fixed and $|b| < 1$, then $n^kb^n \longrightarrow 0$ when $n\longrightarrow \infty$ Show that if $k \in \mathbb{N}$ is fixed and $|b| < 1$, then $n^kb^n \longrightarrow 0$ when $n\longrightarrow \infty$
I have been struggling with this prove. Any suggestions would be great!
 A: You may apply the ratio test:
\begin{align*}
\limsup_{n\rightarrow\infty}\left|\frac{(n+1)^{k}b^{n+1}}{n^{k}b^{n}}\right| = \limsup_{n\rightarrow\infty}\left(\frac{n+1}{n}\right)^{k}|b| = \limsup_{n\rightarrow\infty}\left(1 + \frac{1}{n}\right)^{k}|b| = |b| < 1
\end{align*}
Consequently, the corresponding series
\begin{align*}
\sum_{n=1}^{\infty}a_{n} = \sum_{n=1}^{\infty}n^{k}b^{n} = b + 2^{k}b^{2} + 3^{k}b^{3} + \ldots
\end{align*}
converges. Hence we conclude that $a_{n}\to 0$ just as desired.
A: This is a method using the binomial formula only. We may assume that $0<b<1$. Accordingly $b=(1+\varepsilon)^{-1}$ with $\varepsilon>0$. But $(1+\varepsilon)^n=\sum_{j=0}^n \binom{n}{j}\varepsilon^j>\binom{n}{k+1}\varepsilon^{k+1}$ for $n>k$. Then $0<n^k b^n<\frac{n^k}{\binom{n}{k+1}\varepsilon^{k+1}}\leq c (\frac{n}{n-k})^k \frac1{n-k}$ with some positive constant $c$. But the last expression tends to $0$ for $n\to\infty$ implying the desired result.

A: Let
$$c = \frac{1}{b} \tag{1}\label{eq1A}$$
Since $|b| \lt 1$, you have $|c| \gt 1$. You are thus trying to determine
$$\lim_{n \to \infty}\frac{n^k}{c^n} \tag{2}\label{eq2A}$$
The numerator and the absolute value of the denominator both go to $\infty$. As Representation's question comment indicates, use L'Hôpital's rule $k$ times on the numerator and denominator to get the limit is $0$ since the numerator will be the fixed $k!$ then, but the absolute value of the denominator will still be going to $\infty$.
