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

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

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

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