Show that $f_n(\cdot):[0,1]\rightarrow \Bbb R$, $f_n(x)=x^n(1-x^n)$ is simple convergent, but not uniform convergent.

Question

Not sure if I solved this correctly.

Show that $$f_n(\cdot):[0,1]\rightarrow \Bbb R$$ $$f_n(x)=x^n(1-x^n)$$
is simple convergent, but not uniform convergent.

How I solved it:

$(1)$ $\forall x \in \Bbb R$, $\exists \lim_{n\to \infty}x^n(1-x^n)=0$ $\to$ the series is simple convergent;

$(2)$(I just chose a number in [0,1] that does not converge to 0)
If $x=\frac 12$, $\lim_{n\to \infty}(\frac 12)^n(1-(\frac 12)^n)=-\infty $ $\to$ the series is not uniform convergent;
I think (1) is correct but I am not sure about (2).

Answer 1

^{Answer from comments}

In (1) you prove that the series is simple (pointwise) convergent, in the (2) you prove that the series is NOT simple (pointwise) convergent. Do you realize the mistake in the limit ? -- T_O

For part (2), calculate $f_n((1/2)^{1/n})= (1/2)(1-1/2)=1/4$. -- David Mitra

Answer 2

$\lim_{n\to \infty}(\frac 12)^n(1-(\frac 12)^n)$ is not tending to $-\infty$ because if you take a closer look $(\frac 12)^n$ is tending to $0$ as $\lim_{n\to \infty}$ and so does$(1-(\frac 12)^n)$. So basically the limit tends to $0$.