Welcome to our community

Be a part of something great, join today!

Prove

Alexmahone

Active member
Jan 26, 2012
268
Prove: $\displaystyle \lim\left|\frac{a_{n+1}}{a_n}\right|=L\implies \lim |a_n|^{1/n}=L$.
 
Last edited:

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
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

Active member
Jan 26, 2012
268

Alexmahone

Active member
Jan 26, 2012
268

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
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
Jan 26, 2012
268
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.
 
Last edited:

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
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.