Limit at infinity for sequence $ n^2x(1-x^2)^n$

Question

I'm supposed to prove that this sequence goes to zero as n goes to infinity.

$$\lim_{n\to \infty} {n^2x (1-x^2)^n}, \mathrm{where~} 0 \le x \le 1$$

I've been trying a few things (geometric formula, rewriting $(1-x^2)^n$ as $\sum_{k=0}^n \binom{n}{k} (-1)^k x^{2k} $) and messing around with that. But I can't seem to get anywhere. I could be missing something key that I've forgotten. Can somebody point me in the right direction?

Answer 1

If $x=0$ or $x=1$ it is trivial, so $0<x<1$. Define $a=1-x^2$, then $0<a<1$.

$$\lim\limits_{n\to\infty}n^2x(1-x^2)^n = x\lim\limits_{n\to\infty}n^2a^n \le \lim\limits_{n\to\infty}n^2a^n = \lim\limits_{n\to\infty}\frac{n^2}{(1/a)^n} = \lim\limits_{n\to\infty}\frac{2n}{\ln(1/a)(1/a)^n}=\lim\limits_{n\to\infty}\frac{2}{\ln^2(1/a)(1/a)^n}=\frac{2}{\ln^2(1/a)}\lim\limits_{n\to\infty}a^n=0$$

Using l'Hopital twice and the fact that $0 < a < 1$.

Also all of the terms in the sequence are positive, $0\le\lim\limits_{n\to\infty}n^2x(1-x^2)^n\le 0$.

Answer 2

$$ f_n(x) = x(1-x^2)^n $$ is a positive function on $(0,1)$. By computing $f_n'$, it is easy to check that $x=\frac{1}{\sqrt{2n+1}}$ is the only stationary point of $f_n(x)$ over $(0,1)$, hence: $$ 0 \leq f_n(x) \leq f_n\left(\frac{1}{\sqrt{2n+1}}\right) = \frac{1}{\sqrt{2n+1}}\left(1-\frac{1}{2n+1}\right)^n\leq \frac{1}{\sqrt{2en}}.$$ Moreover, $f_n(x)$ is exponentially small on the interval $\left[\frac{1}{n^{1/3}},1\right]$.

By putting all together, we have that $n^2 f_n(x)$ is a sequence of functions that pointwise converge to the zero function on $[0,1]$. Pointwise but not uniformly, also because: $$ \int_{0}^{1} n^2 f_n(x)\,dx = \frac{n^2}{2n+2}\to +\infty.$$

Answer 3

Hmm, it seems that there was an edit to the question so that I 9 years ago accidentally have answered what was not the final question.

This is now instead an answer to show that $x(1-x^2)^n$ will be 0 as $n\to\infty$

and not for $n^2x(1-x^2)^n$

For an answer to the question as it currently is stated I will recommend you to look at Juan or Jacks answers for that question which look sound. The Rudin theorem in Juan's answer can be phrased in engineering math terms : "exponentials grow faster than potences". Therefore I would probably personally have taken the Rudin route for that problem.

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First let us observe that $x(1-x^2)^n$ is smaller than $(1-x^2)^n$ on the interval, since $|x|<1$.

The largest point of $(1-x^2)$ is at $x=0$ which is $1$ and the smallest is $0$ at $x=1$ (easy to check) and monotonically decreasing. So the function $(1-x^2)^n$ will be bounded by 1. Any $f(x_0) = (1-{x_0}^2)^n$ will decrease exponentially with $n$ for any $x_0\neq 0$ so the crossing $y=\epsilon>0$ will be pushed towards 0 with increasing $n$. So we see that the function $(1-x^2)^n$ will shrink towards $0$ for all values on the interval, except for $x=0$, but as our function of interest is actually $x$ times the function we investigated, so it will of course be even more suppressed.

Answer 4

If $x=0$ or $x=1$ we are done so you can assume that $1>x>0.$ Then $0<1-x^2<1.$

Now put $p:=\dfrac{x^2}{1-x^2}.$ For $n\geqslant4$ we have $$ \begin{aligned} (1+p)^n&>\binom{n}{3}p^3\\\\&=\dfrac{n(n-1)(n-2)}{3!}p^3\\\\&>\dfrac{(n/2)(n/2)(n/2)}{3!}p^3\\\\&=\dfrac{n^3\cdot p^3}{48} \end{aligned} $$ and hence $n^2x(1-x^2)^n=\dfrac{n^2x}{(1+p)^n}\;<\;\dfrac{48x}{p^3}\cdot\dfrac{1}{n}$ and since $\dfrac{48x}{p^3}\cdot\dfrac{1}{n}\to0$ as $n\to\infty$ then $$\lim_{n\to\infty}n^2x(1-x^2)^n=0.$$

In general, you can prove that given any real number $\alpha$ and any real $a\in(0,1)$ then $\lim\limits_{n\to\infty}n^\alpha a^n=0$ using a similar argument.

Answer 5

Yet another approach is as follows. Let $z=\left|\log(1-x^2)\right|$. Then, we have for $x(1-x) \ne 0$

$$\begin{align} n^2x(1-x^2)^{n}&=n^2xe^{n\log(1-x^2)}\\\\ &=\frac{n^2x}{e{nz}}\\\\ &=\frac{n^2x}{1+nz+\frac12 n^2z^2+\frac16 n^3z^3+O(n^4)} \end{align}$$

which clearly approaches $0$ as $n \to \infty$. And we are done as the limit is $0$ for $x=0$ and $x=1$.