MHB Power Series (Which test can i use to determine divergence at the end points)

Click For Summary
The discussion focuses on determining the convergence or divergence of a power series at its endpoints. The user successfully found the interval of convergence for the series f(-4x) = 1/(1+4x) as -1/4 < x < 1/4. Upon testing the endpoint x = -1/4, the series simplifies to a divergent series, confirmed by the divergence test since the limit as k approaches infinity does not equal zero. The conversation clarifies that while the geometric series test is valid, it should specify divergence when r is equal to or greater than 1. The user receives confirmation that their reasoning is correct and valid for determining divergence.
yeny
Messages
7
Reaction score
0
Hello,

I was given f(-4x)= 1/(1+4x), and I used the geometric series to find the power series representation of this function. I then took the limit of (-4x)^k by using ratio test. The answer is abs. value of x. So -1/4<x<1/4

I then plugged in those end points to the series going from k=0 to infinity of (-4x)^k

here's where I'm stuck. How do I determine convergence/divergence of the endpoints?

When I tested x=-1/4, my series was k=0 to infinity of (1)^k, for that series, I wrote " Divergent by divergence test because lim as k --> infinity does not equal zero.

Is that an acceptabe answer? I also had another possible answer which was, Divergent by geometric series because r is less than or equal to 1"

Thank you so much for taking the time to look at this. Hope you all have a wonderful weekend =)
 
Physics news on Phys.org
"When I tested x=-1/4, my series was k=0 to infinity of (1)^k, for that series, I wrote " Divergent by divergence test because lim as k --> infinity does not equal zero."
Yes, that is completely valid

"Is that an acceptabe answer? I also had another possible answer which was, Divergent by geometric series because r is less than or equal to 1"
First, a geometric series is convergent for r< 1. Did you mean "divergent because r is larger than or equal to 1"? I would see no reason to include the "larger than". You are specifically talking about r= 1.

Or, simply, the partial sums are S_n= \sum_{k= 1}^n 1^k= n. What is the limit of that as n goes to infinity.
 
Last edited:
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 17 ·
Replies
17
Views
5K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
3
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K