Simple limit of sequences question.

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Homework Help Overview

The discussion revolves around the convergence or divergence of the sequence defined by {An} = 2^n / n!. Participants are exploring the behavior of this sequence as n approaches infinity.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are questioning the convergence of the sequence and discussing methods to analyze it, including dividing terms and considering the growth rates of the numerator and denominator.

Discussion Status

Some participants have provided suggestions for approaches to analyze the sequence, while others express confusion about the implications of different forms of the sequence. There is an ongoing exploration of the problem without a clear consensus.

Contextual Notes

There are indications of misunderstanding regarding the appropriate forum for such questions, and some participants note the importance of showing effort in the discussion.

nothingkwt
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Let {An} = 2^n / n!

is it convergent or divergent and why?
 
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Not right forum.
No effort shown.
 
Any suggestions yourself?
 
nothingkwt said:
Let {An} = 2n / n!

is it convergent or divergent and why?

(try using the X2 and X2 buttons just above the Rreply box :wink:)

tell us what you think, and why, and then we'll comment! :smile:

(and please use the homework forum in future)
 
Sorry I am new to this forum.

I thought of dividing both terms with the term that approached infinity more rapidly which was n!. which gave me 0, but then I saw that if 2n was actually 10n for example it would approach infinity more rapidly which would give me ∞ but I'm not entirely sure. So I got confused.
 
nothingkwt said:
I thought of dividing both terms with the term …

try dividing one term by the next term instead :smile:

(ie An/An+1)
 
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2*2*2*2*2... / n(n-1)(n-2)(n-3)...

So the limit goes to 0 it's clear now thanks.
 

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