Uniform convergence of the series $\sum_{n=1}^\infty \frac{1}{n} (\frac{x-1}{x+1})^n$

Question

I'm studying the series $$\sum_{n=1}^\infty \frac{1}{n} \cdot (\frac{x-1}{x+1})^n$$

I have already shown that there is pointwise convergence for $x\in[0, +\infty)$ and total convergence in any interval $[0,a]$ with $a>0$, and there is not total convergence in the interval $[0,\infty)$.

How can I prove or disprove the uniform convergence in the interval $[0,\infty)$?

For example: I know that if the series $\sum_{n=1}^\infty f_n(x)$ converges uniformly in $A\subseteq \mathbb{R}$, then the sequence $f_n(x)$ converges uniformly to $0$ in $A$. (This proposition is used to disprove the uniform convergence in a set). But in my case, this test is passed, since $$\text{sup}_{x\in [0,\infty)} |f_n(x)|=1/n$$ whch tends to $0$ as $n$ goes to $\infty$.

In general, are there any other useful tests like the one above to disprove the uniform convergence? Are there sufficient conditions for uniform convergence (other than "total convergence implies uniform convergence")? Which are the the most useful tools (propositions, theorems, etc.) in studying the convergence of a series of functions?

Answer 1

The series fails to converge uniformly for $x \in [1,\infty)$, since for $x > 1$

$$\sup_{x \in [1,\infty)} \left|\sum_{n+1}^\infty \frac{1}{k} \left(\frac{x-1}{x+1} \right)^k\right| \geqslant \sup_{x \in [1,\infty)} \sum_{n+1}^{2n} \frac{1}{k} \left(\frac{x-1}{x+1} \right)^k \geqslant \sup_{x \in [1,\infty)}n \cdot\frac{1}{2n}\left(\frac{x-1}{x+1} \right)^{2n} \\ \geqslant \frac{1}{2} \left(\frac{4n-1-1}{4n -1 +1} \right)^{2n} = \frac{1}{2} \left(1 - \frac{1}{2n}\right)^{2n},$$

and the RHS converges to $e^{-1}/2 \neq 0$ as $n \to \infty$.

Failure to converge uniformly on $[1,\infty)$ implies that convergence in non-uniform on $[0, \infty)$.

Answer 2

I think the following approach is pretty universal to prove that something doesn't converge uniformly on some unbounded interval.

Suppose that $\sum\limits_{n=1}^\infty\frac{1}{n}\left(\frac{x-1}{x+1}\right)^n$ converges uniformly on $[0,\infty)$. Then $\forall\varepsilon>0$ $\exists N\in\mathbb{N}$ such that $\forall x>0$ and $\forall k,m\geq N$ $$|\sum\limits_{n=k}^m\frac{1}{n}\left(\frac{x-1}{x+1}\right)^n|<\varepsilon.$$ Note that $\lim\limits_{x\to\infty}\frac{1}{n}\left(\frac{x-1}{x+1}\right)^n=\frac{1}{n}$ and since $\sum\limits_{n=1}^\infty\frac{1}{n}$ diverges we have $\sum\limits_{n=k}^m\frac{1}{n}\geq\varepsilon$. That means $\exists x_0\in[0,\infty)$ such that $$|\sum\limits_{n=k}^m\frac{1}{n}\left(\frac{x_0-1}{x_0+1}\right)^n|\geq\varepsilon,$$ that is a contradiction. So, $\sum\limits_{n=1}^\infty\frac{1}{n}\left(\frac{x-1}{x+1}\right)^n$ doesn't converge uniformly on $[0,\infty)$.