Is there a series of functions $\sum (-1)^n u_n(x)$ that is convergent for Leibniz rule (alternating series test) in some interval but it is not uniformly convergent?

In particular the series must be (for using Leibniz rule)

  • positive $(u_n(x) \geq 0)$
  • decreasing $(u_{n+1}(x)\leq u_n(x))$
  • such that $\lim_{n \to \infty} u_n(x)=0$

Edit: I refer to the following definition: a series of function is uniformly convergent in a subset $S$ if

$$\lim_{n \to \infty} \sup_{x \in S} |\sum_{n \geq 0} u_n(x) -\sum_{n=0}^{N} u_n(x)| =0$$

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    $\begingroup$ Which rule are you referring to, specifically? (I am used to Leibniz's rule being a name for the alternating series criterion) $\endgroup$ – Clement C. Dec 6 '16 at 17:53
  • $\begingroup$ It's $u_n(x),$ not $u(x)_n.$ Please edit. $\endgroup$ – zhw. Dec 6 '16 at 17:58
  • $\begingroup$ @ClementC. I'm guessing he means $\sum (-1)^nu_n(x)$ $\endgroup$ – zhw. Dec 6 '16 at 18:03
  • $\begingroup$ Wow, that's quite a far cry from what's written. $\endgroup$ – Clement C. Dec 6 '16 at 18:04

Let $$u_n(x)=\frac{x^n}{n}, \quad n\in\mathbb N.$$ Then $u_n: [0,1]\to\mathbb R$ is a decreasing sequence of non-negative functions.

Clearly $$ \sum_{n=1}^\infty u_n(1)=\sum_{n=1}^\infty\frac{1}{n}=\infty, $$ while $$ \sum_{n=0}^\infty (-1)^nu_n(x)=\sum_{n=0}^\infty \frac{(-1)^nx^n}{n}=-\log(1+x), \quad\text{for all $x\in [0,1]$.} $$ In fact, $$ \sum_{k=0}^n (-1)^ku_k(x)=-\log(1+x)-\frac{(-1)^{n+1}\xi^{n+1}}{n+1}, \quad \text{for some $\xi\in (0,x)$}, $$ and thus $$ \bigg|\sum_{k=0}^n (-1)^ku_k(x)+\log(1+x)\,\bigg|\le\frac{1}{n+1}, \quad \text{for all $x\in [0,1]$}, $$ i.e., the series $\,\sum (-1)^nu_n\,$ converges uniformly in $[0,1]$.

Note. That the advantage of this example is that $[0,1]$ is compact.


Let $u_n(x)=e^{-nx}$ where $x\in I=(0,1]$ then

$$R_n(x)=\sum_{k=n+1}^\infty (-1)^k u_k(x)=\frac{(-1)^{n+1} e^{-(n+1)x}}{1+e^{-x}}$$ so $$\Vert R_n\Vert_\infty=\sup_{x\in I}\vert R_n(x)\vert\ge \frac12$$ so the convergence isn't uniform on $I$

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    $\begingroup$ We could also use $u_n(x) = x^n$ on $[0,1)$ - perhaps a bit more elementary. $\endgroup$ – zhw. Dec 9 '16 at 19:39
  • $\begingroup$ is you result true for any decreasing $u_n(x)$ with respect to $x$? $\endgroup$ – DuFong Apr 19 '18 at 3:52
  • $\begingroup$ If $u_n$ is decreasing then we have $$\left\vert\sum_{k=n+1}^\infty (-1)^ku_k(x)\right\vert\le \vert u_{n+1}(x)\vert$$ and in the case where $sup\vert u_n(x)\vert$ tends to 0 we use the above inequality to prove the uniform convergence of the series. @DuFong $\endgroup$ – user296113 Apr 20 '18 at 19:04

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