Prove that $\lim_{n\rightarrow\infty}\sqrt[n]{a_n}=L$

Question

Suppose that $a_n>0$, $n\geq1$ and that $\displaystyle\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}=L$. Prove $\displaystyle\lim_{n\rightarrow\infty}\sqrt[n]{a_n}=L$

To resolve this problem, I solved this one

Suppose that $a_n>0$, $n\geq1$ and that $\displaystyle\lim_{n\rightarrow\infty}a_n=L$. Prove that $\displaystyle\lim_{n\rightarrow\infty}\sqrt[n]{a_1\cdots a_n}=L$.

Here is my progess

$$\lim \sqrt[n]{a_n}= \lim \sqrt[n]{\frac{a_n}{a_{n-1}}\frac{a_{n-1}}{a_{n-2}}\cdots \frac{a_2}{a_1}a_1}=\lim \sqrt[n]{\frac{a_n}{a_{n-1}}\frac{a_{n-1}}{a_{n-2}}\cdots \frac{a_2}{a_1}}\lim\sqrt[n]{a_1}$$

Let $b_n$ be a sequence defined by $$b_n=\frac{a_{n+1}}{a_n}$$

By hypotesis, $\lim b_n=L$. The exercise above leads to

$$\lim \sqrt[n]{a_n}= L\lim\sqrt[n]{a_1}$$.

But what about $\lim\sqrt[n]{a_1}$?

See also: Limit of ${a_n}^{1/n}$ is equal to $\lim_{n\to\infty} a_{n+1}/a_n$. — Martin Sleziak, Jun 20 '24 at 08:07
It is not clear how you conclude that $$\lim\sqrt[n]{\frac{a_n}{a_{n-1}}\frac{a_{n-1}}{a_{n-2}}\cdots \frac{a_2}{a_1}}=L.$$ — user, Jun 20 '24 at 08:24

score 2 · Accepted Answer · answered Mar 30 '20 at 01:49

2

Hint:

For any fixed $c \gt 0$, you have

$$\lim_{x \to 0} c^x = 1 \tag{1}\label{eq1A}$$

answered Mar 30 '20 at 01:49

John Omielan

52,653

Then I put $\lim a_1^{1/n}$, change $1/n$ for $x$ and $n\rightarrow\infty$ to $x\rightarrow0+$. Therefore $\lim a_1^x=a_1^0=1$. – Marcos Paulo Mar 30 '20 at 01:57
@MarcosPaulo Yes, that's correct to prove what you asked about at the end of your post. – John Omielan Mar 30 '20 at 01:59

score 0 · Answer 2 · answered Jun 20 '24 at 08:54

Firstly, we conside the case that $L>0$. Let $0<\alpha<1<\beta$ be arbitrary. Since $\frac{a_{n+1}}{a_{n}}\rightarrow L$, there exists $N$ such that $\alpha L<\frac{a_{n+1}}{a_{n}}<\beta L$ whenever $n\geq N$. By induction, for $k\geq1$, $a_{N+k}<a_{N}\beta^{k}L^{k}$. It follows that \begin{eqnarray*} \sqrt[N+k]{a_{N+k}} & \leq & \sqrt[N+k]{a_{N}}(\beta L)^{\frac{k}{N+k}}. \end{eqnarray*} Letting $k\rightarrow\infty$, we conclude that $\limsup_{n\rightarrow\infty}\sqrt[n]{a_{n}}\leq\lim_{k\rightarrow\infty}\sqrt[N+k]{a_{N}}(\beta L)^{\frac{k}{N+k}}=\beta L$. That is, $\limsup_{n\rightarrow\infty}\sqrt[n]{a_{n}}\leq\beta L$. (Note that $N$ has been discharged.) Since $1<\beta$ is arbitrary, we further conclude that $\limsup_{n\rightarrow\infty}\sqrt[n]{a_{n}}\leq L$.

Similarly, we can prove that $\liminf_{n\rightarrow\infty}\sqrt[n]{a_{n}}\geq\alpha L$, from which we can infer that $\liminf_{n\rightarrow\infty}\sqrt[n]{a_{n}}\geq L$ because $0<\alpha<1$ is arbitrary. Combining the results, we have that $\liminf_{n}\sqrt[n]{a_{n}}=\limsup_{n}\sqrt[n]{a_{n}}=L$, so $\lim_{n}\sqrt[n]{a_{n}}=L$.

Prove that $\lim_{n\rightarrow\infty}\sqrt[n]{a_n}=L$

2 Answers2