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#### Alexmahone

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- Jan 26, 2012

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Prove: $\displaystyle \lim\left|\frac{a_{n+1}}{a_n}\right|=L\implies \lim |a_n|^{1/n}=L$.

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- Thread starter
- #1

- Jan 26, 2012

- 268

Prove: $\displaystyle \lim\left|\frac{a_{n+1}}{a_n}\right|=L\implies \lim |a_n|^{1/n}=L$.

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- Jan 29, 2012

- 661

Denote $q_n=\left|\dfrac{a_{n+1}}{a_n}\right|$. The sequence of positive terms $|q_n|$ converges to $L$, so the sequence if its geometric means also converges to $L$. As $\sqrt[n]{q_0q_1\ldots q_{n-1}}=\sqrt[n]{\dfrac{|a_n|}{|a_0|}}=\dfrac{\sqrt[n]{|a_n|}}{\sqrt[n]{|a_0|}}$, clearly $\sqrt[n]{|a_n|}\to L$ because $\sqrt[n]{|a_0|}\to 1$.Prove: $\displaystyle \lim\left|\frac{a_{n+1}}{a_n}\right|=L\implies \lim |a_n|^{1/n}=L$.

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- Jan 26, 2012

- 268

Why is this true?The sequence of positive terms $|q_n|$ converges to $L$, so the sequence if its geometric means also converges to $L$.

- Jan 29, 2012

- 661

It is a well known result. Look (for example) here:Why is this true?

http://www.sosmath.com/calculus/sequence/hardlim/hardlim.html

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- Jan 26, 2012

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What if $|q_n|=0$ for some $n$?The sequence of positive terms $|q_n|$ converges to $L$, so the sequence if its geometric means also converges to $L$.

- Jan 29, 2012

- 661

We are supposing by hypothesis that $a_{k+1}/a_k$ exists for all $k$. If $q_n=0$ then $a_{n+1}=0$ and this would imply $a_{n+2}/a_{n+1}$ does not exist (contradiction).What if $|q_n|=0$ for some $n$?

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- Jan 26, 2012

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But the only hypothesis is $\displaystyle\lim\left|\frac{a_{n+1}}{a_n}\right|=L$. So, $\displaystyle a_0$ (for instance) could be 0.We are supposing by hypothesis that $a_{k+1}/a_k$ exists for all $k$. If $q_n=0$ then $a_{n+1}=0$ and this would imply $a_{n+2}/a_{n+1}$ does not exist (contradiction).

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- Jan 29, 2012

- 661

Then, $a_1/a_0$ is not defined.But the only hypothesis is $\displaystyle\lim\left|\frac{a_{n+1}}{a_n}\right|=L$. So, $\displaystyle a_0$ (for instance) could be 0.