2
$\begingroup$

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

$\endgroup$
2
$\begingroup$

Hint:

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

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.