Seq Convergence: Does a_n = (-1)^(n+1)/(2n-1) Converge or Diverge?

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The sequence a_n = (-1)^(n+1)/(2n-1) is being analyzed for convergence or divergence. Participants discuss the potential use of the squeeze theorem and the need to evaluate the limit as n approaches infinity. It is suggested that the sequence is conditionally convergent, prompting further questions about the absolute convergence and the relationship between terms a_n and a_{n+1}. Clarification is sought regarding whether the focus is on a sequence or a series, indicating a need for a more complete analysis. The discussion emphasizes the importance of demonstrating the limit to determine the sequence's behavior.
whatlifeforme
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Homework Statement


Converge or diverge?

Homework Equations


\displaystyle a_{n} = \frac{(-1)^{n+1}}{2n-1}

The Attempt at a Solution


all i know is to perhaps take ln of numerator and denominator to get the exponent down below?
 
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whatlifeforme said:

Homework Statement


Converge or diverge?

Homework Equations


\displaystyle a_{n} = \frac{(-1)^{n+1}}{2n-1}

The Attempt at a Solution


all i know is to perhaps take ln of numerator and denominator to get the exponent down below?
Do you know the squeeze theorem ?
 
SammyS said:
Do you know the squeeze theorem ?

I do.
 
whatlifeforme said:
I do.

It appears the sequence is conditionally convergent.

1. What is ##\sum |a_n| = \space ?##

2. Is ##a_n > a_{n+1}##? For sufficiently large n.

3. lim n→∞ ... = ...
 
whatlifeforme said:

Homework Statement


Converge or diverge?

Homework Equations


\displaystyle a_{n} = \frac{(-1)^{n+1}}{2n-1}

The Attempt at a Solution


all i know is to perhaps take ln of numerator and denominator to get the exponent down below?

Are you asking about a sequence or a series? You should fill in part one of the homework template more completely.
 
Last edited:
sequence.
 
So show us what you have done so far in thinking about whether that sequence has a limit.
 

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