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

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?

• What do you mean by "study" simple/uniform convergence? Is the $[0, 1]$ in the title supposed to be $[0, 1[$? – Theo Bendit Sep 30 '18 at 1:37
• by "study" I mean how "to prove " and yes it's supposed to be [0 , 1[ – KEVIN DLL Sep 30 '18 at 1:40
• The first denominator in (1) should be $1-x^2$, not $1-x$. – marty cohen Sep 30 '18 at 1:41
• @martycohen thank , I just edited it – KEVIN DLL Sep 30 '18 at 1:45

## 2 Answers

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.

• thank , but is it possible to show it with Infimum and supremum ? – KEVIN DLL Sep 30 '18 at 2:33
• I'm not sure what you mean. What set would we be taking the infimum or supremum over? Do you have an example (preferably in English) of the type of proof you want? – Theo Bendit Sep 30 '18 at 2:35

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 $$n \to \infty$$.

• can you be more explicit ? – KEVIN DLL Sep 30 '18 at 0:42
• If $x^{2n} \to 0$ uniformly then $x^{2n} <\epsilon$ for $n$ greater then some number $k$, for all $x$. Take $x=1-\frac 1 n$ to get a contradiction when $0<\epsilon <e^{-2}$. – Kavi Rama Murthy Sep 30 '18 at 4:19