# 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

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.
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 $$n \to \infty$$.
• 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