Derivative and integral of a power series proof (not getting a step)

Click For Summary

Homework Help Overview

The discussion revolves around understanding the proof that three specific power series have the same radius of convergence. The original poster is particularly focused on the justification of an inequality used in the proof involving the series and the implications of natural numbers in the context of convergence.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster questions the validity of an inequality in the proof, specifically whether it holds for all natural numbers or only for sufficiently large values of k. Some participants affirm the original poster's concerns and discuss the implications of this on the proof's clarity.

Discussion Status

The discussion is ongoing, with participants exploring the nuances of the inequality and its role in establishing convergence. There is acknowledgment of the original poster's reasoning, and some guidance is offered regarding the sufficiency of the inequality for large k, though no consensus has been reached on the completeness of the proof as presented.

Contextual Notes

Participants note that the proof may lack clarity regarding the conditions under which the inequality holds, suggesting that this oversight could lead to confusion in understanding the proof's validity.

Ryker
Messages
1,080
Reaction score
2

Homework Statement


Basically, there's no problem statement per se, I'm just trying to understand the proof that the following sequences have the same radius of convergence:

[tex](1) \displaystyle\sum_{k=0}^{\infty} c_{k}x^{k}[/tex]
[tex](2) \displaystyle\sum_{k=0}^{\infty} kc_{k}x^{k-1}[/tex]
[tex](3) \displaystyle\sum_{k=0}^{\infty} c_{k}\frac{x^{k+1}}{k+1}[/tex]

The Attempt at a Solution


The proof begins by saying that if (2) converges absolutely, then since [tex]|c_{k}x^{k}| = |x||c_{k}x^{k-1}| \leq |kc_{k}x^{k-1}|[/tex] the comparison test implies the same for (1).

I don't get the inequality part. I mean, since k is a natural number, hence part of an unbounded set, this must be true for all k greater than some k0, but am I missing something that would render this true for all k? I mean, as the proof is stated, nothing suggests that it's supposed to be interpreted as I interpreted it, but if it holds true for all k, then I don't see how this is so.

Any help with understanding this would, as always, be greatly appreciated.
 
Last edited:
Physics news on Phys.org
You're exactly right.
It won't hold for any k, but just for k that are large enough.

However, that is sufficient to determine whether the series converges or not.

This should be mentioned in the proof, so if it's not, that means that they've been sloppy. :wink:
 
Last edited:
I like Serena said:
You're exactly right.
It won't hold for any k, but just for k that are large enough.
Thanks, great to know I was on the right track.
I like Serena said:
However, that is sufficient to determine whether the series converges or not.
Yeah, I know, that's why I first just wanted to let it slide and move on, but after a while decided to check with the PF community anyway, just to see I'm not missing something :smile:
I like Serena said:
This should be mentioned in the proof, so if it's not, that means that they've been sloppy. :wink:
Usually the professor that made these lecture notes would mention something like that, which is what made it all the more puzzling when in this case he didn't. So my first thought was that I'm just not seeing where the step stems from.

I sometimes feel silly making these threads for such small things, but I'm just doing some Maths on my own during summer, and I hate it when I don't get all the steps in a proof or something. I feel I can't just move on, and there's also no one else I could really ask at this time.
 
Glad to have been of help! :smile:
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K