Limit of Sequence: Finding the Limit of a Sequence (2)

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Homework Help Overview

The discussion revolves around finding the limit of the sequence defined by An = n^n / (n+3)^(n+1). Participants are exploring the behavior of this sequence as n approaches infinity.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the initial steps taken in the solution, particularly regarding the use of logarithms and the factoring of n from the denominator. There is also an exploration of rewriting the limit in a different form.

Discussion Status

The discussion is ongoing, with some participants providing corrections and alternative approaches. There is no explicit consensus on the correct method or outcome yet, but guidance has been offered regarding the properties of logarithms and factoring techniques.

Contextual Notes

Some participants express confusion over the manipulation of the sequence and the application of logarithmic properties. There is a mention of a specific limit form that participants are attempting to analyze.

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Homework Statement



Find the limit of this sequence

An =[itex]\frac{n^{n}}{(n+3)^{n+1}}[/itex]

Homework Equations

The Attempt at a Solution



infinite_limit.jpg


The answer is 0 but my answer is 1

thank you
 
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Your first step is wrong.
[tex]\ln \frac{a}{b}≠\frac{\ln a}{\ln b}[/tex]

The correct property is:
[tex]\ln \frac{a}{b}= ln a-\ln b[/tex]

Anyways, instead of taking log on both the sides, you can factor out n from denominator.
 
Pranav-Arora said:
Anyways, instead of taking log on both the sides, you can factor out n from denominator.

Sorry I cannot see how n can factor out from denominator
 
like this ??

http://postimage.org/image/7cqb5oowp/ Thanks
 
Last edited by a moderator:
limit_2.jpg
 
What I actually meant was this:
[tex]\lim_{n→∞} \frac{1}{n+3} \left( \frac{1}{1+\frac{3}{n}} \right)^n[/tex]
 
Thank you
 

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