0

I like to show that $\lim_{n \rightarrow \infty} n^k x^n =0$ if $\vert x \vert <1$ for $k,n \in \mathbb{N}$ and $x \in \mathbb{C}$.

I need the statement to proof a anohter one, i know that for $x=0$ it is trivial but i need a hint to go further.

thanks in advance

McBotto.t
  • 452
sabi
  • 5
  • hint: for $0<\vert x \vert <1$ you can write $\vert x \vert $ as $ \vert x \vert = \frac{1}{1+h}$ for an $h>0$ – McBotto.t Jan 28 '18 at 18:00
  • thank you, so then i have

    $\lim_{n \rightarrow \infty} \vert n^k x^n \vert =\lim_{n \rightarrow \infty} n^k \vert x \vert^n = \lim_{n \rightarrow \infty}n^k (\frac{1}{1+h})^n $

    but now i have some trouble to estimate $(\frac{1}{1+h})^n$ if this is the idea?

     – sabi Jan 28 '18 at 18:16
  • what is $k$ here ? – the_firehawk Jan 28 '18 at 18:18
  • k is a natural number – sabi Jan 28 '18 at 18:21
  • The idea is to estimate $(1+h)^n > ... > \lambda n^{k+1}$ with one $\lambda > 0$ by using binomial theorem. – McBotto.t Jan 28 '18 at 18:24
  • Related: https://math.stackexchange.com/questions/55468/how-to-prove-that-exponential-grows-faster-than-polynomial – Hans Lundmark Jan 29 '18 at 09:18

2 Answers2

3

only when $x$ is real you have the following :

if $x = 0$ it's trivial

if $x > 0 $

$n^kx^n = e^{k\ln n + n\ln x} = e^{k\ln n -cn}, \; \; c > 0$

since ${k\ln n -cn} \to - \infty$ then $n^kx^n \to 0$

if $x < 0$

$- n^k(-x)^n \leq n^kx^n = (-1)^n n^k(-x)^n \leq n^k(-x)^n $

by the squeeze theorem you get what you want to prove

if $x \in \mathbb{C}, \; \; x=re^{i\theta}$ since $0 < r < 1$ and $e^{i\theta}$ is bounded you have the same result as if $x$ was real

the_firehawk
  • 2,425
1

If the limit exists, then, by continuity, $$ \left|\lim_{n\rightarrow\infty}n^kx^n\right|=\lim_{n\rightarrow\infty}n^k|x|^n. $$ Moreover, if the RHS is $0$, then the LHS exists and is also $0$. Let's look at ratios of successive pairs. In particular, $$ \frac{(n+1)^k|x|^{n+1}}{n^k|x|^n}=\left(\frac{n+1}{n}\right)^k|x|. $$

For $n$ sufficiently large, $\left(\frac{n+1}{n}\right)^k|x|$ can be bounded away from $1$ as follows: Suppose that $|x|=1-\varepsilon$. Since $\frac{n+1}{n}$ is decreasing to $1$, we may find an $N$ such that if $n\geq N$, $\left(\frac{n+1}{n}\right)^k<1+\varepsilon$. In particular, we need $$ \frac{n+1}{n}\leq (1+\varepsilon)^{1/k}. $$ In other words, $$ n\geq \frac{1}{(1+\varepsilon)^{1/k}-1}. $$ With this choice of $N$, $\left(\frac{n+1}{n}\right)^k|x|<1-\varepsilon^2<1$. Let $$ y=1-\varepsilon^2. $$ By construction, we know that for $n\geq N$, $$ \frac{(n+1)^k|x|^{n+1}}{n^k|x|^n}\leq y $$ or that $$ (n+1)^k|x|^{n+1}\leq y n^k|x|^n. $$ By induction, we get that for $n\geq N$, $$ n^k|x|^n\leq N^k|x|^Ny^{n-N}. $$ Therefore, we consider $$ \lim_{n\rightarrow\infty}N^k|x|^Ny^{n-N}=N^k\left(\frac{|x|}{y}\right)^N\lim_{n\rightarrow\infty}y^n. $$ This equals $0$ since $0<y<1$. Since $0\leq n^k|x|^n\leq N^k|x|^Ny^{n-N}$ for $n$ sufficiently large, by the squeeze theorem, we get that the original limit is $0$.

Michael Burr
  • 32,867