I need to proof convergence, and find the radius of converge

In summary, the conversation is about two homework problems, one involving proving convergence and the other finding the radius of convergence. The participants discuss using the comparison test and choosing a specific value for y in order to justify the convergence. They also talk about the first factor in the second problem being a power series with a limit of e. The conversation ends with a reminder to show why 1/(a_n-2) converges and a hint to use the comparison test.
  • #1
asi123
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0

Homework Statement



Hey guys, can you help me with this on please?
First one, I need to proof convergence, and the second one is to find the radius of converge.

Homework Equations





The Attempt at a Solution

 

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  • #2
For the first one, you are given the first series converges. That means a_n->infinity. Think comparison test. 1/(a_n-x)<1/(a_n-y) if x<y<a_n. How about choosing y=a_n/2?? Can you justify that? For the second one, write it as [(n+1)^n/n^n]*[1/n^z]. The first factor has a limit. What is it?
 
  • #3
Dick said:
For the first one, you are given the first series converges. That means a_n->infinity. Think comparison test. 1/(a_n-x)<1/(a_n-y) if x<y<a_n. How about choosing y=a_n/2?? Can you justify that? For the second one, write it as [(n+1)^n/n^n]*[1/n^z]. The first factor has a limit. What is it?

Right, it's e. so that's mean that in order for the series to converge, x need to bigger then 1?
 
  • #4
Looks to me like aside from the e, it's a power series.
 
Last edited:
  • #5
Dick said:
Looks to me like aside from the e, it's a power series.

This are my thoughts, is this right?
 

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  • #6
asi123 said:
This are my thoughts, is this right?

Why do you think 1/(a_n-2) converges? Shouldn't you state a reason?
 
  • #7
Dick said:
Why do you think 1/(a_n-2) converges? Shouldn't you state a reason?

If 1/(a_n-2) converges, than why shouldn't 1/(a_n-2) converge? I mean if a_n -> infinity than I don't think that 2 will bother him, no?
 
  • #8
asi123 said:
If 1/(a_n-2) converges, than why shouldn't 1/(a_n-2) converge? I mean if a_n -> infinity than I don't think that 2 will bother him, no?

No, the 2 won't bother him. But you still have to show that. Set up a comparison test with something you know converges. Review my hint about this one.
 
  • #9
Dick said:
No, the 2 won't bother him. But you still have to show that. Set up a comparison test with something you know converges. Review my hint about this one.

Yeah, I got you, thanks.
 

What is convergence?

Convergence refers to the behavior of a sequence or series as its terms approach a specific value or limit. In other words, it is the tendency of a sequence or series to get closer and closer to a certain value as more terms are added.

What is the difference between convergence and divergence?

Convergence and divergence are opposite behaviors of a sequence or series. Convergence occurs when the terms of a sequence or series approach a specific value or limit, while divergence occurs when the terms do not approach a specific value or limit and instead increase or decrease without bound.

How do you prove convergence?

To prove convergence, you can use various methods such as the squeeze theorem, the monotone convergence theorem, or the ratio test. These methods involve analyzing the behavior of the sequence or series in question and determining if it approaches a specific value or limit.

What is the radius of convergence?

The radius of convergence is a measure of how quickly a power series approaches a specific value or limit. It is the distance from the center of the series to the point where the series converges. The larger the radius of convergence, the faster the series converges to its limit.

How do you find the radius of convergence?

The radius of convergence can be found by using the ratio test or the root test, which determine the values of x for which the series will converge. Additionally, the radius of convergence can also be found by using the Taylor series expansion of a function, which gives the interval of convergence for that specific function.

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