Uniform convergence of $f(x) = \sum_{n=0}^\infty x^n(1-x)^n$ in $(0,1)$

Question

How to show uniform convergence of $$f(x) = \sum_{n=0}^\infty x^n(1-x)^n$$ for $x$ in $(0,1)$ ?

I know I have to find the convergence radius but I don't know how.

Answer 1

Hint Note that $$ 0\leq\sum_{n=0}^m x^n(1-x)^n\le \sum_{n=0}^m \frac{1}{4^n} $$ and use the fact that $\sum_0^\infty 4^{-n}$ converges. Now let $m\to \infty$.

Answer 2

Radius of convergence is given by $|x(1-x)|=1$ or $x^2-x-1=0$. Solution is $x=\frac{1\pm \sqrt{5}}{2}$. Use smaller magnitude value to get $r=\frac{\sqrt{5}-1}{2}$.

Answer 3

Let us use calculus to find the maximun of $f$ in $[0,1]$ :

Here $f(x)=x^n(1-x)^n$. Then $$f'(x)=n(x-x^2)^{n-1}(1-2x)$$

Now $f'(x)=0$ if $x-x^2=0$ or $1-2x=0$, which is happen if $x=0$ or $1$ or $1/2$

That is, $0,1,1/2$ are critical points

Here $f(0)=0=f(1)$ and $f(1/2)=\frac{1}{4^n}$

Hence the maximum of $f $ is $\frac{1}{4^n}$.

That is, $$f(x) \leq \frac{1}{4^n}$$ for all $x \in [0,1]$

Now use Foobaz John hint to conclude your result!