MHB Approaching Infinity - Finding the Limit of a Sequence

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The limit of the sequence is evaluated as $\lim_{n\to\infty}\frac{\sqrt[n]{n}}{\sqrt[n+1]{n+1}}$. It is established that $\sqrt[n]{n} = e^{\frac{\ln n}{n}}$, and as $n$ approaches infinity, $\frac{\ln n}{n}$ approaches 0, leading to $\sqrt[n]{n}$ converging to 1. The limit $\lim_{u \to \infty} \sqrt[u]{u}$ is also confirmed to be 1. Consequently, the final answer to the original problem is 1.
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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