# Prove

#### Alexmahone

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

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#### Fernando Revilla

##### Well-known member
MHB Math Helper
Prove: $\displaystyle \lim\left|\frac{a_{n+1}}{a_n}\right|=L\implies \lim |a_n|^{1/n}=L$.
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 $\sqrt[n]{|a_0|}\to 1$.

• Alexmahone

#### Alexmahone

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

#### Fernando Revilla

##### Well-known member
MHB Math Helper
• Alexmahone

#### Alexmahone

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

#### Fernando Revilla

##### Well-known member
MHB Math Helper
What if $|q_n|=0$ for some $n$?
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).

#### Alexmahone

##### Active member
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).
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.

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#### Fernando Revilla

##### Well-known member
MHB Math Helper
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.
Then, $a_1/a_0$ is not defined.