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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?

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  • $\begingroup$ what is total convergence? $\endgroup$
    – zhw.
    Aug 17, 2019 at 16:12
  • $\begingroup$ @zhw, I have heard how some people (with Italian background) referred to uniform convergence as a "total convergence" and think that OP is among them. $\endgroup$
    – Hasek
    Aug 17, 2019 at 16:14
  • $\begingroup$ @zhw. By total convergence I mean normal convergence. (In Italy we call it in this way) :) $\endgroup$
    – Minato
    Aug 17, 2019 at 16:38
  • $\begingroup$ math.stackexchange.com/questions/1356517/… $\endgroup$ Aug 17, 2019 at 18:23

2 Answers 2

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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 $[0, \infty)$.

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  • $\begingroup$ Wonderful and very elegant! Can you give some hints on how did you come up with this calculations? It seems so magical...I could never have thought of such a thing $\endgroup$
    – Minato
    Aug 17, 2019 at 22:14
  • $\begingroup$ @Minato: Once you establish that the series converges pointwise, i.e., the sequence of partial sums $S_n(x)$ converges to $S(x)$, then uniform convergence for $x \in D$ means for any $\epsilon > 0$ there exists $N$ independent of $x$ such that $|S_n(x) - S(x)| < \epsilon$ for all $n > N$ and all $x \in D$. This is true if and only if $\sup_{x \in D} |S_n(x) - S(x)| \to 0$. The sequence of remainders $|S_n(x) - S(x)|$ is $\left|\sum_{k = n+1}^\infty a_k(x)\right|$ and I showed in this case it can't converge to $0$ with a lower bounding sequence that fails to converge to $0$. $\endgroup$
    – RRL
    Aug 17, 2019 at 22:27
  • $\begingroup$ Bounding below by a sum with finitely many terms, $\sum_{k=n+1}^{2n} a_k(x)$ is a common trick. It is also equivalent to showing that a uniform Cauchy criterion is not satisfied. $\endgroup$
    – RRL
    Aug 17, 2019 at 22:31
  • $\begingroup$ There are many examples like this on Math.SE, for example my answer here. $\endgroup$
    – RRL
    Aug 17, 2019 at 22:34
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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 $[0,\infty)$.

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  • $\begingroup$ Thank you. Please let me rephrase your answer to see if I understood it well. Suppose the series $\sum\limits_{n=1}^\infty\frac{1}{n}\left(\frac{x-1}{x+1}\right)^n$ converges uniformly in $[0,\infty)$. Let $\varepsilon>0$, then there exists $\nu \in \mathbb{N}$ such that for all $m\ \ge k \ge \nu$ we have $$\left|\sum\limits_{n=k}^m\frac{1}{n}\left(\frac{x-1}{x+1}\right)^n\right| <\varepsilon/2 \text{ for all $x$ in $[0,\infty)$}$$ $\endgroup$
    – Minato
    Aug 17, 2019 at 22:04
  • $\begingroup$ Taking the sup for $x\in [0,\infty)$ we have $$\left|\sum\limits_{n=k}^m\frac{1}{n}\right| \le\varepsilon/2<\varepsilon $$ So we have shown that for every $\varepsilon >0$ there is $\nu \in \mathbb{N}$ such that for every $m \ge k \ge \nu$ we have $$\left|\sum\limits_{n=k}^m\frac{1}{n}\right| <\varepsilon$$, i.e. that the series $\sum\limits_{n=1}^\infty\frac{1}{n}$ converges, which is absurd! Am I right? $\endgroup$
    – Minato
    Aug 17, 2019 at 22:23
  • $\begingroup$ @Minato, yes, you're right. I also corrected a typo with quantifiers order in my answer. $\endgroup$
    – Hasek
    Aug 18, 2019 at 0:02

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