# Examine the uniform convergence of the series of functions $f_n: \mathbb{R} \to \mathbb{R}: x \mapsto \frac{(-1)^n}{n}$

Examine the uniform convergence of the series of functions $f_n: \mathbb{R} \to \mathbb{R}: x \mapsto \frac{(-1)^n}{n}$

Attempt:

Note that for any $x \in \mathbb{R}$:

$$\sum_n^\infty f_n(x) = \sum_n^\infty \frac{(-1)^n}{n}$$

and by the alternating series criterium (Leibniz), the series $\sum_{n}^\infty f_n(x)$ converges for every $x \in X$. Hence, the series of functions $\sum_n^\infty f_n$ converges pointwise. Because the series does not depend on $x$, it also converges uniformly.

Is this correct?

• It looks fine...but I find this question a little odd. You're given functions $\;f_n(x):=\frac{(-1)^n}n\;$ ...so these are constant functions... Dec 26, 2017 at 10:29
• Yes this is the first in a whole bunch of exercises. The following ones depend on $x$.
– user370967
Dec 26, 2017 at 10:31
• I will post a few and if my methods are fine then I'll be confident.
– user370967
Dec 26, 2017 at 10:31
• I've no idea, but after more than 15 months in this site you should be used to that. Somebody just didn't like what you wrote for some mysterious, and perhaps even stupid, reason, so (s)e downvoted you. I upvoted it at once as I think your work shows a nice effort and some insight for understanding things you're studying now. And don't mind about downvotes: they shall come and go for no special reasons many times. Dec 26, 2017 at 15:17
• Yeah. Guess I will have to get used to that. Thanks for your comment.
– user370967
Dec 26, 2017 at 15:19

Yes. The terms follow all the conditions of the Leibniz alternating series convergence theorem. The terms alternate in sign, decrease in magnitude, and tend to zero as $$n$$ approaches infinity. Since $$(t_n = s_n - s_{n-1})$$, we can say that $$\sum_{n}^\infty f_n(x)$$ does not depend on $$x$$.