Derivative of (-1)^n/l'hopital's rule

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Homework Help Overview

The discussion revolves around demonstrating the convergence of a series using the alternating series test. The series in question is given by the expression \((-1)^n/(2\sqrt{3n})\), and participants are exploring the necessary conditions for convergence, particularly focusing on the limit of the absolute value of the terms.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need to show that the limit of the absolute value of the terms approaches zero and whether the sequence is decreasing. There is confusion regarding the application of L'Hôpital's rule to the expression involving \((-1)^n\), with some questioning its relevance.

Discussion Status

Several participants have provided insights into the requirements of the alternating series test, emphasizing the importance of the limit and the monotonicity of the sequence. There is ongoing clarification about the role of \((-1)^n\) in the limit process, and some participants suggest reviewing notes or external resources for further understanding.

Contextual Notes

There is mention of potential confusion regarding the application of L'Hôpital's rule, as well as the nature of the series' sum, which cannot be expressed in elementary terms. Participants are also addressing the monotonicity condition for the sequence involved in the test.

vinnie
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Homework Statement



I am tryint to show that a series converges using the alternating series test. First, we show that 0<an+1<an and then that the limit of the expression equals 0. (the question says show that the series converges using the alternating series test, so I know it converges.)

Homework Equations



l'hopital's rule,

The Attempt at a Solution



the series is ((-1)^n)/(2sqrt(3n))

I need to use l'hopital's to show that the limit equals zero. The problem is, what is the derivative of (-1)^n if ln (-1) doesn't exist?
 
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Hi Vinnie,

If a_n is the absolute value of the nth term (i.e., the factor being multiplied by (-1)^n), in this case 1/(2*sqrt(3n)), then part of the alternating series test asks that \lim_{n\to \infty} a_n = 0, so (-1)^n does not occur in the limit.
 
so we would take the limit of 1/2srqrt(3n)?

how would we find the sum of this infinite series?
 
vinnie said:
so we would take the limit of 1/2srqrt(3n)?
For the second part of alternating series test, yes. The other part of the test, as you have said, requires you to show that the sequence 1/(2*sqrt(3n)) (a_n as I defined previously) is decreasing. Review your notes.

how would we find the sum of this infinite series?
The alternating series test tells you the series converges, which is all the question required. This particular series' sum cannot be written in elementary terms. In mathematics, determining precisely what a quantity is may be a completely different task to simply showing that the quantity exists.
 
Step 1: Determine a_n
Step 2: Determine if the SEQUENCE a_n is monotone decreasing

ie: The limit is 0 and it is infact decreasing a_n < a_(n-1)

Wikipedia should have more info on the alternating series test if this is not clear enough.

Note: Not sure why you want to use l'hospital, its not applicable.
 
moo5003 said:
Step 1: Determine a_n
Step 2: Determine if the SEQUENCE a_n is monotone decreasing

ie: The limit is 0 and it is infact decreasing a_n < a_(n-1)

Wikipedia should have more info on the alternating series test if this is not clear enough.

Note: Not sure why you want to use l'hospital, its not applicable.

Now that I know about (-1)^n, I know it's not applicable.
 
Unco said:
For the second part of alternating series test, yes. The other part of the test, as you have said, requires you to show that the sequence 1/(2*sqrt(3n)) (a_n as I defined previously) is decreasing. Review your notes.


The alternating series test tells you the series converges, which is all the question required. This particular series' sum cannot be written in elementary terms. In mathematics, determining precisely what a quantity is may be a completely different task to simply showing that the quantity exists.

I was speaking of the second part of the test. I know all about the first part. Thank you.
 
To show 1/(2sqrt(3n)) is decreasing, you could take the derivative of 1/(2sqrt(3x)) and show that it is negative for positive x, or just note that since n is increasing 1/sqrt(n) must be decreasing.
 

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