Approaching Infinity - Finding the Limit of a Sequence

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Discussion Overview

The discussion centers around finding the limit of the sequence defined by $\displaystyle\lim_{n\to\infty}\frac{\sqrt[n]{n}}{\sqrt[n+1]{n+1}}$. Participants explore the behavior of the sequence as \( n \) approaches infinity, involving mathematical reasoning and limit evaluation.

Discussion Character

  • Mathematical reasoning, Debate/contested

Main Points Raised

  • Some participants express that $\sqrt[n]{n}=e^{\frac{\ln n}{n}}$ and note that $\frac{\ln n}{n}\to 0$ as \( n\to\infty\).
  • Others propose that this leads to the conclusion that the limit is $\displaystyle\frac{1}{1}=1$.
  • Several participants reiterate the limit $\displaystyle\lim _{u \to \infty } \sqrt[u]{u}=1$ as part of their reasoning.
  • There are multiple affirmations of the limit being 1, but the original question remains open to interpretation.

Areas of Agreement / Disagreement

While some participants agree that the limit is 1, the discussion does not reach a consensus on the final answer to the original post, as it is framed as a question rather than a statement of fact.

Contextual Notes

The discussion includes repeated references to the limit of $\sqrt[n]{n}$ and its behavior, but does not resolve the implications of these evaluations fully. There is also some ambiguity regarding the interpretation of the original limit posed.

Who May Find This Useful

Readers interested in mathematical limits, sequences, and the behavior of functions as they approach infinity may find this discussion relevant.

alexmahone
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Find $\displaystyle\lim_{n\to\infty}\frac{\sqrt[n]{n}}{\sqrt[n+1]{n+1}}$.
 
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$\sqrt[n]{n}=e^{\frac{\ln n}{n}}$ and $\frac{\ln n}{n}\to 0$ as $n\to\infty$.
 
Evgeny.Makarov said:
$\sqrt[n]{n}=e^{\frac{\ln n}{n}}$ and $\frac{\ln n}{n}\to 0$ as $n\to\infty$.

So are you saying that the answer is $\displaystyle\frac{1}{1}=1$?
 
Alexmahone said:
Find $\displaystyle\lim_{n\to\infty}\frac{\sqrt[n]{n}}{\sqrt[n+1]{n+1}}$.
Do you know this limit
$\displaystyle\lim _{u \to \infty } \sqrt{u}=~?$
 
Plato said:
Do you know this limit
$\displaystyle\lim _{u \to \infty } \sqrt{u}=~?$


It's 1.
 
Alexmahone said:
It's 1.
What is the answer to the OP?
 
Plato said:
What is the answer to the OP?

$\displaystyle\frac{1}{1}=1$
 
I agree.
 

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