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Let $a_n$ be a positive sequence Which of the following statements is true?

  1. if $\lim \limits_{n \to \infty} \sqrt[n]{a_n}>1$ then $\lim \limits_{n \to \infty} a_n = \infty$
  2. if $a_n$ converges or diverges to infinity then $\sqrt[n]a_n$ converges
  3. if $\lim \limits_{n \to \infty}{a_n}= \infty$ then $\lim \limits_{n \to \infty} \sqrt[n]{a_n}>1$
  4. if $\sqrt[n]a_n$ converges then $a_n$ converges or diverges to infinity

I believe the correct answer is $1$ because I found contradicitions to the rest but could not prove the first

for the second statement it is not true because let $ a_n= \begin{cases} 2^n&\text{if}\, n _{even}\\ 3^n&\text{if}\, n_{odd}\\ \end{cases} $ then for all $n$ we get $\lim _\limits {n \to \infty} a_n= \infty $ but $\lim \limits_{n \to \infty} \sqrt[n]{a_{2n}}=2$ and $\lim \limits_{n \to \infty} \sqrt[n]{a_{2n-1}}=3$ so the limit does not exist

for the third statement let $a_n=n$ then $\lim \limits_{n \to \infty} {a_n}= \infty$ and $\lim \limits_{n \to \infty} \sqrt[n]{a_n}=1$

and for the fourth statement let $ a_n= \begin{cases} n&\text{if}\, n _{even}\\ 1&\text{if}\, n_{odd}\\ \end{cases} $ so $\lim \limits_{n \to \infty} \sqrt[n]{a_{2n}}=1=\lim \limits_{n \to \infty} \sqrt[n]{a_{2n-1}}=1$ but the limit for $a_n$ does not exist because $\lim \limits_{n \to \infty} {a_{2n}}= \infty$ and $\lim \limits_{n \to \infty} {a_{2n-1}}=1$

But I could not prove or begin with the correct statement which is the first ( according to what I did it is the correct one)

Thanks for any tips and help!

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    $\begingroup$ For 1., you find $\epsilon>0$ such that $\lim_n\sqrt[n]{a_n}>1+\epsilon$. Hence $a_n>(1+\epsilon)^n$ for almost all $n$. The right member of the inequality diverges to $\infty$, hence $a_n$ as well. $\endgroup$
    – Zuy
    May 22, 2022 at 12:49
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    $\begingroup$ if the limit exists and it is $c>1$, then by definition of limit $a_n>(c-\epsilon)^n$ definitively $\endgroup$
    – Exodd
    May 22, 2022 at 12:49

3 Answers 3

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Let $L=\lim_n \sqrt[n]{a_n}$ Then $L>1$. Fix $1<c<L$. THen for some $N\in\mathbb{N}$, $\sqrt[n]{a_n}>c$ for all $n\geq N$. That means that $$a_n\geq c^n,\qquad n\geq N$$ Can you finish from here?

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Assume $\sqrt[n]{a_n}\to L > 1$. Then for $\varepsilon := \frac{L-1}2 > 0$ there exists $n_0 \in \Bbb{N}$ such that $$n \ge n_0 \implies \sqrt[n]{a_n} \ge L-\varepsilon = \frac{L+1}2.$$ In particular for all $n \ge n_0$ we have $$a_n \ge \left(\frac{L+1}2\right)^n \xrightarrow{n\to\infty} +\infty$$ because $\frac{L+1}2 > 1$.

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The first is correct. Let $c > 1$ be the limit. Fix $\epsilon > 0$, then by definition, there exists $N$ such that $\forall n > N, |\sqrt[n]{a_n} - c| < \epsilon \implies a_n > (c - \epsilon)^n$. In particular, we take $\epsilon = \frac{c - 1}{2}$, then $c' := c - \epsilon > 1$ and

$$ a_n > (c - \epsilon)^n = c'^n \stackrel{n \to \infty}{\to} \infty $$


For the fourth one, I believe $a_n = 1$ works just as well.

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