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I am studying whether the function-series $\sum_{n=1}^\infty f_n$with

$$f_n=(-1)^n\frac{x}{n}$$ converges pointwise, absolutely, uniformly on $$E=[-1, 1]\rightarrow Y=\mathbb{R}.$$

a) pointwise convergence:
$\sum_{n=1}^\infty f_n$ converges for every x in $E$ because of leibniz-criterium. Therefore $\sum_{n=1}^\infty f_n$ converges pointwise.

b) absolute convergence
$\sum_{n=1}^\infty f_n$ doesn't converge absolutely because $|(-1)^n\frac{x}{n}|$ results in the harmonice series $\sum_{n=1}^\infty \frac{1}{n}x$ which doesn't converge.

c) uniform convergence
In order to converge uniformly

$$\lim_{n\rightarrow\infty} sup_{x\in[-1,1]}|f_n(x)-f(x)|<\epsilon$$ where $f(x)$ is $ln(2)$ and $sup_{x\in[-1,1]}|f_n(x)-f(x)|$ is $ln(2) -f_1(1)=ln(2)+1$, which does not converge to $0<\epsilon$. Therefore we got no uniform convergence.

I would appreciate it if you could point out any mistakes. I am especially not sure if the argumentation for c) holds. Thank you!

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a) Correct.

b) Almost correct. After all, the series converges absolutely when $x=0$.

c) Totally wrong. First of all, $f(x)=-\log(2)x$, not $\log(2)$. And in order to prove that the convergence is not uniform, what you ought to prove is that$$\lim_{n\to\infty}\sup_{x\in[-1,1]}\left|f(x)-\sum_{k=1}^nf_k(x)\right|=0.$$In this case, it's only a matter of noticing that, if $a_n=\left|-\log(2)-\sum_{k=1}^n\frac{(-1)^k}k\right|$, then $\lim_{n\to\infty}|a_n|=0$ and that\begin{align}\sup_{x\in[-1,1]}\left|f(x)-\sum_{k=1}^nf_k(x)\right|&=\sup_{x\in[-1,1]}\left|-\log(2)x-\sum_{k=1}^n\frac{(-1)^k}kx\right|\\&\leqslant|a_n|.|x|.\end{align}

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  • $\begingroup$ Thank you! Can you show me how to do c)? $\endgroup$ – lolan496 Feb 10 '18 at 12:32
  • $\begingroup$ I've edited My answer. I hope that it is clear now. $\endgroup$ – José Carlos Santos Feb 10 '18 at 12:44

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