Proving $\lim_{x\to \infty} a_n = 0$ if $\lim_{x\to \infty} \sqrt[n]{|a_n|} = a$ for $a < 1$

Question

I received an assignment to prove this by definition of limits.
$$ \lim_{x\to \infty} \sqrt[n]{|a_n|} = a ; a < 1 \Rightarrow\lim_{x\to \infty} a_n = 0 $$

The prove should analogically work for: $a > 1$

I have honestly no idea how to even begin to prove it. Could anybody lend me a hand? I'd be more than grateful. Limits are something that just strongly denies to get understood by me.

Answer 1

You can construct the following inequality rather easily: $$\forall\epsilon>0; \sqrt[n]{|a_n|}<a+\epsilon$$ this is true because it is given that $\sqrt[n]{|a_n|}=a$ so adding any positive number to the right side will create that I inequality.

Further more for large enough $n$ I can achive the following:$$\exists\epsilon>0,\ \epsilon+a<1; \sqrt[n]{|a_n|}<a+\epsilon$$this is true because if $\sqrt[n]{|a_n|}=a,a<1$ I can add to the right side a number as small as I like.

Using the last inequality I get: $$\exists\epsilon>0,\ \epsilon+a<1; \sqrt[n]{|a_n|}<a+\epsilon\implies {|a_n|}<(a+\epsilon)^n\\\text{let's take the limit:}\\\lim_{n\to\infty}{|a_n|}<(a+\epsilon)^n$$ because $a+\epsilon<1$ the limit as n goes to infinity of $(a+\epsilon)^n$ goes to $0$ and because $|a_n|\ge0$ I can use the squeeze theorem I have $\lim_{n\to\infty}|a_n|=0\implies\lim_{n\to\infty}a_n=0$

Answer 2

Let $a<b<1$ then for large enough $n$, $$\sqrt[n]{|a_n|}<b$$ thus $$|a_n|<b^n$$ now use that $b^n\to 0$.

Answer 3

We have that for every $\epsilon > 0$ there is an $N \in \mathbb{N}$ such that for every $n > N$, $| \sqrt[n]{|a_n|} - a | < \epsilon$.

Hence, $0 \le \sqrt[n]{|a_n|} < a + \epsilon$. Suppose that $\epsilon$ is small enough that $a+ \epsilon < 1$.

Then we have that $0 \le |a_n| < (a+ \epsilon)^n$ for all $n > N$. What do we know happens to $r^n$ when $0 < r < 1$ ?