Some help with finding the limit of a series

  • Thread starter Thread starter vande060
  • Start date Start date
  • Tags Tags
    Limit Series
Click For Summary

Homework Help Overview

The discussion revolves around determining the convergence or divergence of a sequence defined as an = ((-1)n-1n)/(n2 + 1). Participants are exploring the behavior of this sequence as n approaches infinity.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants express uncertainty about how to begin analyzing the sequence and mention the potential use of L'Hospital's rule. There is a discussion about the relationship between the numerator and denominator, with one participant suggesting that the denominator grows faster than the numerator.

Discussion Status

Some participants have pointed out the distinction between sequences and series, which has prompted clarification. There is a suggestion to manipulate the sequence for easier analysis, and one participant expresses a growing understanding of the problem. The limit of the absolute value of the sequence is noted to be zero, leading to further discussion about the implications for the original sequence.

Contextual Notes

Participants are navigating initial confusion regarding terminology and the methods applicable to the problem. The original poster's uncertainty about the techniques learned in class is evident, and there is a focus on foundational concepts related to limits and convergence.

vande060
Messages
180
Reaction score
0

Homework Statement



determine whether or not the sequence converges of diverges. If it converges, find the limit.

an = ((-1)n-1n)/(n2 + 1)

Homework Equations

I have a feeling this theorem is used

if lim n->∞ |an| = 0, then lim n->∞ an = 0

The Attempt at a Solution



I don't really know where to start on this one, I thought about using L'Hospital for this, but I don't know how to do exponential derivatives without taking a log, and I don't remember the teacher ever doing that in class when looking at this variety of problem
 
Physics news on Phys.org
vande060 said:

Homework Statement



determine whether or not the sequence converges of diverges. If it converges, find the limit.

an = ((-1)n-1n)/(n2 + 1)



Homework Equations

I have a feeling this theorem is used

if lim n->∞ |an| = 0, then lim n->∞ an = 0



The Attempt at a Solution



I don't really know where to start on this one, I thought about using L'Hospital for this, but I don't know how to do exponential derivatives without taking a log, and I don't remember the teacher ever doing that in class when looking at this variety of problem

Do you have any sense of what this one does?
bn = n/(n2 + 1)

For n > 0, the denominator is always larger than the numerator, and gets larger more rapidly.
 
Mark44 said:
Do you have any sense of what this one does?
bn = n/(n2 + 1)

For n > 0, the denominator is always larger than the numerator, and gets larger more rapidly.

yes i understand that, that would mean that the series is decreasing right?
 
You seem to be using the terms "series" and "sequence" interchangeable, but they are different. Your thread title includes the word series, but what you showed is a sequence.

For your sequence, it might help to pull a factor of n2 out of the numerator, like so:
[tex]a_n = (-1)^{n - 1}\frac{n}{n^2(1 + 1/n^2)}[/tex]

This really isn't a very complicated sequence.
 
Mark44 said:
You seem to be using the terms "series" and "sequence" interchangeable, but they are different. Your thread title includes the word series, but what you showed is a sequence.

For your sequence, it might help to pull a factor of n2 out of the numerator, like so:
[tex]a_n = (-1)^{n - 1}\frac{n}{n^2(1 + 1/n^2)}[/tex]

This really isn't a very complicated sequence.

Im sorry, I am sure its not complicated, but this is one of the first ones I've seen:blushing:

I think I understand how to do the rest now though.

If I find the limit of the absolute value of this sequence I find that it is zero. By the theorem I posted earlier, the limit of the original sequence with the -1 is also 0. Correct?

Thanks for your help so far
 
Yes, the limit of the absolute values of this sequence is zero, which means that the limit of the original sequence is also zero.
 

Similar threads

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