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Sagot :
The given sequence [tex]2^{\frac{1}{n} }[/tex] converges with limit 1.
According to the given question.
We have a sequence,
[tex]a_{n} = 2^{\frac{1}{n} }[/tex]
Since, we know that
The sequence of real numbers [tex]S_{n}[/tex], where n goes from 1 to infinity has a limit L, the the sequence is convergent to L. If the sequence doesn't have limit then it is divergent.
The nth root function is strictly increasing for positive real values.
Therefore,
[tex]1^{\frac{1}{n} } \leq 2^{\frac{1}{n} } \leq n^{\frac{1}{n} }[/tex] [tex]\forall \ n\geq 2[/tex]
Also,
[tex]\lim_{n \to \infty} n^{\frac{1}{n} } = 1[/tex]
[tex]\implies \lim_{n \to \infty} 1^{\frac{1}{n} } =1[/tex] ( by squeeze theorem)
Thereofore,
[tex]\lim_{n \to \infty}2^{\frac{1}{n} } =1[/tex]
So, the given sequecence has a limit i.e. 1. Which means [tex]2^{\frac{1}{n} }[/tex] is a convergent sequence.
Hence, the given sequence [tex]2^{\frac{1}{n} }[/tex] converges with limit 1.
Find out more information about convergent and divergent sequence here:
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