limit of sequence of functions $f_n(x)=nx(1-x^2)^n$

Question

I have function defined on $x\in[0,1]$ $$f_n(x)=nx(1-x^2)^n$$ I need to find its limit function. $$\lim_{n\to\infty}f_n(x)=f(x) \\ $$ Through plotting the function I can determine that $f(x)=0:x\in(0,1]$ I am not able to prove this analytically.

Answer 1

It is clear that $f(0) = f(1) = 0$.

In a very rare cases, a test, such as ratio test or root test etc., for series can deduce whether a sequence converges to $0$ or not.

In this case, root test can be used.

For $ 0 < x < 1$, remember that by root test the series $$\sum_{n = 1}^\infty f_n(x)$$
converges if $$C :=\limsup_{n\to \infty} (f_n)^{1/n} < 1 $$ But $C = 1 - x^2 < 1$, which mean the series converge. On the other hand, we know that a series converges implies its nth term tend to zero. Thus $$\lim_{n\to\infty} f_n(x) = 0$$ for $0 < x < 1$.

Answer 2

Indeed, you are correct. Let $x \in (0, 1]$. Then by the continuity of the logarithm we say if $$L \overset{\text{def}}{=} \lim_{n \to \infty} n (1-x^2)^n$$ then $$L = \exp \left( \lim_{n \to \infty }\log (n) + n \log(1-x^2)\right)$$ For our fixed $x$ we can pick $n$ sufficiently large so that we have $\log n \leq -\frac{1}{2}n \log(1-x^2) $, whence for $n$ sufficiently large we have $$ \frac{1}{2}n \log(1-x^2) \geq \log (n) + n \log(1-x^2) $$ and thus we can make the argument of our exponential as small as we like (arbitrarily negative, that is). Thus $L=0$.

Answer 3

So the problem is, $\displaystyle\lim_{n\to \infty}nt^n=0$ for all $t\in [0,1)$. Note that, $t=\frac{1}{1+r}$ for some $r>0$. For any $\epsilon>0$ and any positive integer $n>\frac{2}{\epsilon r^2}+1$ we have, $$nt^n=\frac{n}{(1+r)^n}=\frac{n}{1+nr+\frac{n(n-1)}{2}r^2+\cdots+r^n}\leq\frac{n}{\frac{n(n-1)}{2}r^2}=\frac{2}{(n-1)r^2}<\epsilon.$$ So we are done.