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

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 $$\lim\sqrt[n]{a_1}$$?

For any fixed $$c \gt 0$$, you have
$$\lim_{x \to 0} c^x = 1 \tag{1}\label{eq1A}$$
• 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 at 1:57