Does the series of functions, with $ f_n(x) = (-1)^n x^n (1-x)$ converge uniformly on $[0,1]$

Question

I have tried reducing the problem into $g_n(x) = (-1)^n x^n$, as the sum of two uniformly convergent series on some $C$ is uniformly convergent on $C$.

Answer 1

Notice that:

$f_n(x)$ converges pointwise to $0$

$|f'_n(x)|=nx^{n-1}(1-x)-x^n$ then $\max |f_n(x)|$ is when $x=\frac{n}{n+1}$

So, $\forall x \in [0,1]$, $|f_n(x)-f(x)|=|f_n(x)| \le |f_n(\frac{n}{n+1})|=\frac{n^n}{(n+1)^n}\frac{1}{n+1} \le \frac{1}{n+1} \rightarrow0$

And waht can we say about $\sum_{n=0}^\infty (-1)^nx^n(1-x)$?

Notice that $|\sum_{n=0}^j(-1)^n| \le 1 \forall j \in \mathbb{N}$

We have already proved that $ x^n(1-x) \rightarrow 0$ uniformly

Now, observe that the sequence $x^n(1-x)$ is non-increasing $\forall x \in [0,1]$ and $\forall n \in \mathbb{N}$

Then we can use Dirichlet's Test for Uniform Convergence(page 34) and the series $\sum_{n=0}^\infty (-1)^nx^n(1-x)$ converges unifomly in $[0,1]$

Answer 2

Using a mean inequality we get an inequality holding for $x\in [0,1]$: $$0\le x^n\cdot (1-x)=n^n\cdot \frac{x\cdot x\cdot ...\cdot x}{n\cdot n\cdot ...\cdot n}\cdot (1-x)\le n^n\cdot \left(\frac{n\cdot \frac xn+(1-x)}{n+1}\right)^{n+1}=\frac {n^n}{(n+1)^{n+1}}<\frac{1}{n+1}.$$ Let $o: [0,1]\to \mathbb R$ be $o(x)=0$. Then: $$||f_n-o||=\sup_{x\in [0,1]} |(-1)^n x^n(1-x)|<\frac1{n+1},$$ so it can be smaller than any $\varepsilon>0$ by dropping enough $f_n$'s and your sequence converges uniformly to $o$.

Edit: I was asked to explain the application of mean equality goes as follows: we use $n+1$ non-negative numbers: $(1-x)$ and $n$ numbers $x/n$. Then:

$$\sqrt[n+1]{(1-x)\cdot\frac{x^n}{n^n}}\le \frac{(1-x)+n\cdot \frac xn}{n+1}=\frac 1{n+1}$$

Answer 3

The series $\sum_{n=1}^{\infty} (-1)^{n}x^{n}(1-x)$ does converge uniformly on $[0,1]$: The partial sums are given by $S_n(x)=\frac {1-(-x)^{n+1}} {1+x} (1-x)$. To show that this expression converges uniformly to $\frac {1-x} {1+x}$ we have to show that $\frac {(-x)^{n+1}} {1+x} (1-x) \to 0$ uniformly. Consider $x \in [0,1-\delta]$ and $x \in [1-\delta,1]$ separately. Given $\epsilon >0$ there exists $\delta$ such that $|\frac {(-x)^{n+1}} {1+x} (1-x)|\leq \frac {1-x} {1+x} <\epsilon $ for $x \in [1-\delta,1]$. For $x \in [0,1-\delta]$ we have $|\frac {(-x)^{n+1}} {1+x} (1-x)| \leq {(1-\delta)^{n+1}}$ which $\to 0$.

Answer 4

Let $\,g_n(x)=|\,f_n(x)|=x^n(1-x)$. Then $$ g_n'(x)=n(1-x)x^{n-1}-x^n=x^{n-1}\big(n(1-x)-x\big)=x^{n-1}\big(n-(n+1)x\big) $$ and $\,g'_n(x)=0\,$ if and only if $\,x=\frac{n}{n+1}$. The maximum hence of $g_n$ is assumed at $\,x=\frac{n}{n+1}$ and hence $$ |\,f_n(x)|\le g_n(x)\le g_n\Big(\frac{n}{n+1}\Big)=\Big(\frac{n}{n+1}\Big)^n\Big(1-\frac{n}{n+1}\Big)=\frac{1}{n+1}\Big(\frac{n}{n+1}\Big)^n<\frac{1}{n+1}, $$ and hence $f_n$ converges uniformly to zero.