Study simple convergence of $\sum_{n=0}^{\infty} x^{2n}$ on $[0,1[$

Question

I have to study

1 ) the simple convergence of

$$S(x) = \sum_{n=0}^{\infty} x^{2n}$$

and

2) the uniform convergence

My attempts :

1) $\forall x \in [0,1[$ $$S_n(x) = \frac{1-(x^2)^{n+1}}{1-x^2}$$

The series converges if and only if $|x| < 1$ so, $\forall x \in [0,1[$

$$S(x) = \sum_{n = 0}^{ \infty} x^{2n} = \frac{1}{1-x^2}$$

Can you check my solution and if there is an error can you help me?

Answer 1

As you said, the $n$th partial sum is $$S_n = \frac{1 - x^{2(n+1)}}{1 - x^2}.$$ Do these partial sums converge uniformly to the limit $\frac{1}{1 - x^2}$? Equivalently, does $$\frac{1}{1 - x^2} - \frac{1 - x^{2(n+1)}}{1 - x^2} = \frac{x^{2(n+1)}}{1 - x^2} \to 0$$ uniformly? If it did, then for any $\varepsilon > 0$, we could find an $N$ such that the following holds for all $x \in [0, 1)$: $$n > N \implies \frac{x^{2(n+1)}}{1 - x^2} < \varepsilon.$$ But, for any fixed $n$, we have $$\lim_{x \to 1^-} \frac{x^{2(n+1)}}{1 - x^2} = \infty.$$ This means, if we fix some $n > N$, then there exists some $\delta > 0$ such that $$1 - \delta < x < 1 \implies \frac{x^{2(n+1)}}{1 - x^2} > \varepsilon,$$ which contradicts uniform convergence.

Answer 2

Your answer for 1) is correct. If the series converges uniformly then $x^{2n} \to 0$ uniformly. But this is false because $(1-\frac 1 n) ^{2n} \to e^{-2}$ s $n \to \infty$.