Question

Determine the convergence or divergence of the sequence with the given nth term. if the sequence converges, find its limit. (if the quantity diverges, enter diverges. ) an = 21/n

Answer 1

The given sequence $2^{\frac{1}{n} }$ converges with limit 1.

According to the given question.

We have a sequence,

$a_{n} = 2^{\frac{1}{n} }$

Since, we know that

The sequence of real numbers $S_{n}$, 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,

$1^{\frac{1}{n} } \leq 2^{\frac{1}{n} } \leq n^{\frac{1}{n} }$           $\forall \ n\geq 2$

Also,

$\lim_{n \to \infty} n^{\frac{1}{n} } = 1$

$\implies \lim_{n \to \infty} 1^{\frac{1}{n} } =1$   ( by squeeze theorem)

Thereofore,

$\lim_{n \to \infty}2^{\frac{1}{n} } =1$

So, the given sequecence has a limit i.e. 1. Which means $2^{\frac{1}{n} }$ is a convergent sequence.

Hence, the given sequence $2^{\frac{1}{n} }$ converges with limit 1.

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