# Let $a_n$ be a positive sequence prove that if $\lim \limits_{n \to \infty} \sqrt[n]{a_n}>1$ then $\lim \limits_{n \to \infty} a_n = \infty$

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!

• 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.
– Zuy
May 22, 2022 at 12:49
• if the limit exists and it is $c>1$, then by definition of limit $a_n>(c-\epsilon)^n$ definitively May 22, 2022 at 12:49

Let $$L=\lim_n \sqrt[n]{a_n}$$ Then $$L>1$$. Fix $$1. THen for some $$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$$ Can you finish from here?
Assume $$\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$$.
The first is correct. Let $$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.