Why does $\sum_{n=0}^{\infty} (-1)^n x^n (1-x)$ converges uniformly on $[0,1]$?

Question

I'm trying to solve problem 9.11 from Apostol's: Mathematical Analysis. Specifically I'm trying to show that: $$ \sum_{k=0}^{\infty} (-1)^k x^k (1-x)$$ converges uniformly on $[0,1]$

I have seen that Dirichlet's test can be used to prove that the series is uniformly convergent in $[0,1]$, but while solving the problem I seemed to find a contradiction. Could someone help me to check if my reasoning is correct?

For each $n \in \mathbb{N}$ let $f_n(x) = \sum_{k=0}^n (-1)^k x^k (1-x)$ so that $f_n(0)=f_n(1)=0$. Hence for each $x \in (0,1)$:

$$ f_n(x) = (1-x) \sum_{k=0}^n (-x)^k = \dfrac{1-x}{1+x} (1-(-x)^{n+1})$$ So that $\lim \limits_{n \to \infty} f_n(x) = \dfrac{1-x}{1+x}$ and therefore the sequence $(f_n)$ converges pointwise to the function:

$$ f(x) = \begin{cases} 0, \text{ if } x=0\\ \dfrac{1-x}{1+x}, \text{ if } 0<x<1\\ 0, \text{ if } x = 1 \end{cases}$$

But $f$ is not continuous on $[0,1]$, Isn´t this a contradiction for uniform convergence? (since each $f_n$ is continuous because it is the finite sum of continuous functions and the uniform limit of continuous functions must be a continuous function)

Answer 1

$$\text{let :}a_n = \sup_{x \in (0,1)} {|f_n(x)|}$$ $\sum a_n \text{converge} \iff \sum f_n(x) \text{converges uniformly}$

Put : $$f_n(x)=(-1)^n x^n (1-x) \implies |f_n(x)|=x^n(1-x) \; \; x \in (0,1)$$

Therfore : $$a_n = \sup_{x \in (0,1)} {|f_n(x)|} =|f_n\left({\frac{n-1}{n}}\right)|=\left({1-\frac{1}{n}}\right)^n\cdot \frac{1}{n}=\frac{(n-1)^n}{n^{n+1}}$$

We have : $$\sum a_n \text{divege} \implies \sum f_n(x) \text{ Not converges uniformly}$$

Answer 2

It is possible to provide a direct proof. On the other hand we can use the Dini's theorem: if $s_n(x)$ is a monotonic sequence of continuous functions on $[0,1],$ convergent pointwise to a continuous function $s(x),$ then the convergence is uniform. In the OP case the pointwise limit of the partial sums $s_n(x)$ is equal $(1-x)/(1+x).$ Moreover $s_{2n}$ is increasing while $s_{2n+1}$ is decreasing. Therefore the convergence is uniform.

The direct proof can be performed as follows. We have $$|s(x)-s_{n-1}(x)|={1-x\over 1+x}x^n$$ For $1\ge x\ge 1-1/\sqrt{n}$ we get $|s(x)-s_n(x)|\le 1/\sqrt{n}$ while for $0\le x\le 1-1/\sqrt{n}$ we get $$|s(x)-s_n(x)|\le \left (1-{1\over \sqrt{n}}\right )^n\\ =\left (1+{1\over \sqrt{n}-1}\right )^{-n}$$ By the Bernoulli inequality we obtain $$\left (1+{1\over \sqrt{n}-1}\right )^{n}\ge 1+{n\over \sqrt{n}-1}\ge \sqrt{n}$$ Hence $$|s(x)-s_n(x)|\le {1\over \sqrt{n}}$$ Remark The Dirichlet test can be used as well, but we would need to show that $x^n(1-x)$ tends to $0$ uniformly, which was the key point in the direct proof. Besides the latter can be proved applying the derivatives and finding that the maximal value is attained at $n/(n+1).$ Hence $x^n(1-x)\le {1\over n+1}.$ I have given up the derivatives in the second proof above to make the solution more elementary.