Question about Uniform Convergence of Series of Functions

Question

I am trying to find if a series of functions is uniform convergent on [-1, 1], or (negative infinite, infinite) or any other interval.

Here is the series:

Here is the partial sum:

I found in the excellent book Real Analysis, by John Howie, an easy way to prove it.

Take the absolute value of the difference between the infinite sum and the partial sum. And analyze what happens when n goes to infinite.

Here is the absolute value of the difference, and to that value we take the limit that is 0.

If I understood correctly, as the limit when n goes to infinite is 0, the series is uniformly convergent for all real numbers.

Here are my questions:

1) Is my answer correct?

2) If I want to use the other uniform convergence definition, find N that depends only on epsilon, how can I do it?

Here is the other definition of uniform convergence that I want to use, written by William Wade

Answer 1

Note that :$$\forall x \in \Bbb R, \quad \frac{1}{x^2+1} <1 \iff x \neq 0 \iff x \in \Bbb R^* $$

That means that you want find if you have uniform convergence on $\Bbb R^*$

If we use $f(x)$ as sum of the serie, we have $$\forall x \in \Bbb R^*, \quad R_n(x)=|f_n(x)-f(x)|=\sum\limits_{k=n+1}^{+\infty}\frac{x^2}{(x^2+1)^k}=\frac{x^2}{(x^2+1)^{n+1}}.\frac{x^2+1}{x^2}=\frac{1}{(x^2+1)^n}$$

Then $$\alpha_n=\sup\limits_{x\in \Bbb R^*}(R_n(x))=1$$ Since the sequence $(\alpha_n)$ do not converge s to $0$ , we have not uniform convergence on $\Bbb R^*$

However, the convergence is uniform on every set of the form $X_a=]-\infty,a] \cup [a,+\infty[,$ where $a$ is a real numbre such that $a > 0$, since : $$\beta_n=\sup\limits_{x\in X_a}(R_n(x))=\frac{1}{(1+a^2)^n}$$ and $$\lim_{n \to +\infty } \beta_n=0$$