How to show that the series of functions $\sum_{k = 0}^\infty(-1)^k \frac{x^{2k + 1}}{2k + 1}$ is pointwise convergent to a function?

Question

The problem I'm given is as follows:

Consider the sequence of functions $(f_k)_k$, where $f_k(x) = (-1)^k \frac{x^{2k + 1}}{2k + 1}$ as well as the resulting series: $$A(x) = \sum_{k = 0}^\infty f_k (x)$$ Show that the series converges pointwise towards a function $f(x)$ for all $x \in (-1, 1)$. It is not yet necessary to show which $f$ the series converges to.

My confusion in answering this question arises from the fact that the tests (Cauchy and M-Test) determine uniform convergence, which is stronger than pointwise. I understand that passing either test would be necessary and sufficient for uniform convergence and therefore also for pointwise, but since the question asks only to show pointwise convergence, I'm a bit lost.

My approach is as follows:

By the ratio test, I have shown that the series converges absolutely (to a value) for all $x \in (-1, 1)$.
The next step should be to somehow justify that the values for each $x$ that the series converges to belong to a function, but I am not sure how to do that.

If anyone could help me to structure the last step, I'd be very grateful.

you can fix $x_0$ and then study the numerical series $\sum f_k(x_0)$ — Sine of the Time, Jan 04 '24 at 19:55
Probably here the usage of Cauchy-Hadamard theorem is mentioned? — Egor Ivanov, Jan 04 '24 at 19:56
@SineoftheTime could you please elaborate on that approach? Power series were very loosely introduced, so not really. — johnjorgenson, Jan 04 '24 at 20:05
using theorems related to power series, you can find the radius of convergence. — Sine of the Time, Jan 04 '24 at 20:06
Does this answer your question? Show that the series $\sum_{k=0} ^\infty (-1)^k \frac{x^{2k+1}}{2k+1}$ converges for $|x|<1$ and that it converges to $\arctan x$ (look at the beginning of the accepted answer) alternatively, $\left|(-1)^k \frac{x^{2k + 1}}{2k + 1}\right|\le|x|^{2k + 1}$ and $\sum_{k=0}^\infty|x^{2k+1}|=\frac{|x|}{1-x^2}<\infty$ hence your series is absolutely convergent. — Anne Bauval, Jan 04 '24 at 20:07
@AnneBauval It's a start... we haven't been introduced to what a radius of convergence is and I was looking for an answer with more explanation — johnjorgenson, Jan 04 '24 at 20:12
In my alternative solution in the previous comment, I did not use the radius of convergence, not even the ratio test. Sorry, I could have edited my solution as an answer but now the post is closed. — Anne Bauval, Jan 04 '24 at 20:12
I understand the use of the ratio test, what I'm not grasping is the relation between absolute and pointwise convergence — johnjorgenson, Jan 04 '24 at 20:14
@PriyamTurakhia if you can't use power series and you want to use the most basic tools, just fix $x_0$ and study the behaviour of $\sum f_k(x_0)$ — Sine of the Time, Jan 04 '24 at 20:15
@AnneBauval it's OK, I reckon I can piece something together with the other post — johnjorgenson, Jan 04 '24 at 20:16
Look at https://en.wikipedia.org/wiki/Absolute_convergence It is much more fundamental than the alternating series test. Absolute convergence of a series of complex numbers (not functions) famously implies ordinary convergence. Both are "pointwise" in your context, i.e. for $x$ fixed. — Anne Bauval, Jan 04 '24 at 20:18

score 0 · Accepted Answer · answered Jan 04 '24 at 20:10

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If we really have to constraint ourselves to the minimal level of math knowledge needed. Since you mentioned no integral, I assume no Taylor series either, so there is no way for you to tell what $f(x)$ is, but you do not seem to need that.

The most direct way is the ratio test, which proves the absolute convergence of the series. May be this suffice?

answered Jan 04 '24 at 20:10

Frank

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But how do absolute and pointwise convergence relate? – johnjorgenson Jan 04 '24 at 20:12
@PriyamTurakhia if the function's value absolutely converges (as a series) at each $x$, then the function pointwise converges. – Frank Jan 04 '24 at 20:17

How to show that the series of functions $\sum_{k = 0}^\infty(-1)^k \frac{x^{2k + 1}}{2k + 1}$ is pointwise convergent to a function?

1 Answers1