What is the Limit of a Sequence of Exponents?

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Homework Help Overview

The discussion revolves around evaluating the limit of a sequence of exponents, specifically the expression lim_{n→∞} \frac{ (n+1)^{\frac{1}{n+1}} }{n^{\frac{1}{n}}}. Participants are exploring the behavior of this limit as n approaches infinity.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Some participants attempt to analyze the limit by considering the convergence of n^{1/n} and suggest that if this limit exists, it could imply that the overall limit of the fraction is 1. Others question the validity of this reasoning, pointing out that the growth of n may affect the limit differently than expected.

Discussion Status

The discussion is ongoing, with participants providing different perspectives on the limit. Some have offered insights into the behavior of the terms involved, while others are questioning the assumptions made about the convergence of the sequence. There is no explicit consensus yet, but the exploration of different interpretations is evident.

Contextual Notes

Participants are navigating the complexities of limits involving sequences and exponents, with some expressing confusion over the application of L'Hopital's rule and the implications of terms approaching zero. The original poster mentions a connection to a series problem and the ratio test, indicating a broader context for the limit being discussed.

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Homework Statement



Evaluate [tex]lim_{n→∞} \frac{ (n+1)^{\frac{1}{n+1}} }{n^{\frac{1}{n}}}[/tex]

Homework Equations


The Attempt at a Solution


This is actually a part of a series problem I am trying to solve using the ratio test.I can't seem to figure out this limit and L'Hopital's doesn't work. Any hints?

BiP
 
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Bipolarity said:

Homework Statement



Evaluate [tex]lim_{n→∞} \frac{ (n+1)^{\frac{1}{n+1}} }{n^{\frac{1}{n}}}[/tex]
Note that if [itex]\lim_{n \rightarrow \infty} n^{1/n}[/itex] exists, let's call it [itex]L[/itex], then both the numerator and denominator converge to [itex]L[/itex], so the limit of the fraction is 1.

Indeed, [itex]\lim_{n \rightarrow \infty} n^{1/n}[/itex] does exist, so focus on proving that fact.
 
hey i think the limit is 1

1/(n+1) → 0
and
1/n → 0

if both powers will go → 0 then everything0 = 1


then you get 1/1 = 1
 
Helpeme said:
hey i think the limit is 1

1/(n+1) → 0
and
1/n → 0

if both powers will go → 0 then everything0 = 1


then you get 1/1 = 1

The answer is right, but the argument is invalid. It's true that [itex]x^{1/n} \rightarrow 1[/itex] if [itex]x[/itex] is a fixed positive number, but here we have [itex]x[/itex] growing to infinity while the exponent shrinks to zero. It is not automatically true that the limit will be 1.

Consider for example
[tex]\lim_{n \rightarrow \infty} (n^n)^{1/n}[/tex]
Surely this does not converge to 1, since [itex](n^n)^{1/n} = n[/itex].
 

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