Consider $\sum\limits_{n=0}^\infty (-1)^nx^{2n}$. Does it converges uniformly in the intervals $-1<x<1$ and $-1\le x\le 1$?

Question

Consider the geometric series $\sum\limits_{n=0}^\infty (-1)^nx^{2n}$. Does it converges uniformly in the intervals $-1<x<1$ and $-1\le x\le 1$?

The answer of the book says no in $-1<x<1.$

• My argument to justify the no convergence in $[-1, 1]$:

We have to show that $\max\limits_{-1\le x\le1} \Big|f(x)-\sum\limits_{n=1}^\infty f_n(x)\Big| \to 0$ when $n\to\infty$ (which is the definition for $[-1,1]$) doesn't hold.

The series has ratio $-x^2$, thus the $n$-th partial sum is $S_n=\frac{1-(-x^2)^{n+1}}{1+x^2}$.

And $\max\limits_{-1\le x\le1}\Big|f(x)-\sum\limits_{n=1}^\infty f_n(x)\Big|=\max\limits_{-1\le x\le1}\Big|\frac{-(-x^2)^{n+1}}{1+x^2}\Big|\not\to 0$ when $n\to\infty$ (justification is $n$-th term test).

Hence does not converge uniformly in $[-1,1]$

• Note that we cannot even consider convergence in $(-1,1)$ since the definition only applies for closed intervals.

Answer 1

For $[-1, 1]$, you're correct. For $x = 1$, the $\max$ does not go to $0$.

However, for $(-1, 1)$, I don't think the definition for uniform convergence requires closed intervals. You can indeed speak meaningfully about uniform convergence on open-intervals, like in this Math.SE question.