Prove that $\lim_{n \to \infty} k^nn^p= 0$ where $|k| < 1$ and $p>0$ Hello I am working through some problems in a book and came across this question.
Prove that $\lim_{n \to \infty} k^nn^p= 0$ where $|k| < 1$ and $p>0$ 
I can see why it should be the case (since exponentials grow faster than polynomials) but I don't really know where to start to try and prove it rigourously. I have identified that it becomes a $"0 \times \infty"$ situation and have tried using L'Hôpital's rule on the expression $k^n/(1/n^p)$ which didn't seem to help. How should go about tackling this problem?
 A: A stronger conclusion is possible using ratio test. $\frac {k^{n+1} (n+1)^{p} } {k^{n}n^{p}} \to k <1$ so the series $\sum k^{n}n^{p}$ converges. This implies that $k^{n}n^{p} \to 0$. 
A: One does not need taking derivatives to prove rigorously the result. 


*

*When $0<k<1$, this is immediate from the following theorem:

If $p>0$ and $\alpha$ is real, then $\displaystyle\lim_{n\to\infty}\frac{n^\alpha}{(1+p)^n}=0$. 

(Note that $0<\frac{1}{1+p}<1$ plays the role of $k$ in your limit.)
The above is a theorem in Rudin's Principle of Mathematical Analysis (Theorem 3.20 (d) on page 57 of the 3rd edition), which can be proved by a few words using binomial theorem.†
In this case, this question was asked before in this site. See this answer to an old popular question of this site: How to prove that exponential grows faster than polynomial?

*When $k=0$, the result is trivial. 

*When $-1<k<0$, you could rewrite the limit as
$$
\lim_{n\to\infty}(-1)^n(-k)^nn^p
$$
and use the result in the first case.

† Proof. Let $k$ be an integer such that $k>\alpha$, $k>0$. For $n>2k$,
$$
(1+p)^n>\binom{n}{k}p^k=\frac{n(n-1)\cdots(n-k+1)}{k!}p^k>\frac{n^kp^k}{2^kk!}.
$$
Hence
$$
0<\frac{n^\alpha}{(1+p)^n}<\frac{2^kk!}{p^k}n^{\alpha-k}\quad (n>2k).
$$
Since $\alpha-k<0$, $n^{\alpha-k}\to 0$. 
A: It suffices to show 
\begin{align*}
(|k|^{1/p})^{n}n\rightarrow 0.
\end{align*}
Let $a=|k|^{1/p}<1$, we are to show that $a^{n}n\rightarrow 0$.
Let $a=1/(1+r)$ for $r>0$, then $a^{n}\leq\dfrac{1}{1+nr+n(n-1)r^{2}/2}$, it is now easy to show that
\begin{align*}
\dfrac{n}{1+nr+n(n-1)r^{2}/2}\rightarrow 0.
\end{align*}
A: 
So I tried applying l'Hôpital's rule $p$ times on the expression $k^n/(1/n^p)$
  . I then get $\frac{k^n ln(k)^p}{(-1)^p*p(p+1)(p+2)...(2p)) / n^{2p}}$ and I don't quite see where that gets me.

l'Hôpital is possible, but if you do it that way
$$ \lim_{n\to\infty} k^n n^p =\lim_{n\to\infty}  \frac{k^n}{n^{-p}}=\lim_{n\to\infty} \frac{k^{n}\log k}{(-p)n^{-p-1}} = \dots$$
you end up with terms of the same structure. If instead you use the $\infty/\infty$ l'Hôpital rule, you can reduce the power of $n^p$, and this lets you conclude. Say $p\notin \mathbb N$, then
\begin{align} \lim_{n\to\infty} k^n n^p 
&=\lim_{n\to\infty}  \frac{n^p}{k^{-n}}
\\&= \lim_{n\to\infty} p\frac{n^{p-1}}{k^{-n
} (-\log k)} 
\\&= \lim_{n\to\infty} p(p-1) \frac{n^{p-2}}{k^{-n
} (-\log k)^2} 
\\&\mathrel{ \vdots}\\
& = \lim_{n\to\infty} p(p-1)\dots(p-\lfloor p \rfloor)(-\log k)^{-\lfloor p \rfloor - 1}n^{p-\lfloor p \rfloor - 1} k^n 
\\
&= p(p-1)\dots(p-\lfloor p \rfloor)(-\log k)^{-\lfloor p \rfloor - 1} \lim_{n\to\infty} n^{p-\lfloor p \rfloor - 1} k^n 
\\
&= p(p-1)\dots(p-\lfloor p \rfloor)(-\log k)^{-\lfloor p \rfloor - 1} \lim_{n\to\infty} n^{p-\lfloor p \rfloor - 1}\lim_{n\to\infty} k^n 
\\&=  p(p-1)\dots(p-\lfloor p \rfloor)(-\log k)^{-\lfloor p \rfloor - 1} 0\times 0
\\ &= 0
\end{align}
the proof is easier if $p\in\mathbb N$.
