TITLE: 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 [0 upvotes]: 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).

REPLY [0 votes]: $\lim_{n\to \infty}(\frac 12)^n(1-(\frac 12)^n)$ 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 $0$.
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