MHB Prove $\lim_{n \to \infty} \sqrt[n]{a_1a_2...a_n} = L$ for Convergent ${a_n}$

[mathgirl1]

Assume the sequence of positive numbers ${a_n}$ converges to L. Prove that
$\lim_{n \to \infty} \sqrt[n]{a_1a_2...a_n} = L$ (The nth root of the product of the first n terms)

Since ${a_n}$ converges we know that for every $\epsilon> 0$ there is an $N$ such that for all $n > N$ $ |a_n - L| < \epsilon$. Need to show that for every $\epsilon > 0$ there is an $N$ s.t. for $n > N$ $|\sqrt[n]{a_1a_2...a_n} - L | < \epsilon$. Not sure how to relate these two. I know that if $a_n$ converges to $L$ then $\sqrt{a_n}$ converges to $\sqrt{L}$. I can extend this to say that $\sqrt[n]{a_n}$ converges to $\sqrt[n]{L}$. I also know the product of two convergent sequences converges to the product of the limit of each sequence. Not sure how to extend that to be the product of the first $n$ terms of a convergent sequence is convergent. Maybe Cauchy criteria is the way to proceed but I didn't have much luck in that either.

Any help would be appreciated! Thanks in advance!

 

[chisigma]

Assume the sequence of positive numbers ${a_n}$ converges to L. Prove that
$lim_{n \to \infty} \sqrt[n]{a_1a_2...a_n} = L$ (The nth root of the product of the first n terms)

Since ${a_n}$ converges we know that for every $\epsilon> 0$ there is an $N$ such that for all $n > N$ $ |a_n - L| < \epsilon$. Need to show that for every $\epsilon > 0$ there is an $N$ s.t. for $n > N$ $|\sqrt[n]{a_1a_2...a_n} - L | < \epsilon$. Not sure how to relate these two. I know that if $a_n$ converges to $L$ then $\sqrt{a_n}$ converges to $\sqrt{L}$. I can extend this to say that $\sqrt[n]{a_n}$ converges to $\sqrt[n]{L}$. I also know the product of two convergent sequences converges to the product of the limit of each sequence. Not sure how to extend that to be the product of the first n terms of a convergent sequence is convergent. Maybe Cauchy criteria is the way to proceed but I didnt have much luck in that either.

Any help would be appreciated! Thanks in advance!

By Cesaro theorem, given a sequence $a_{n}$ and the sequence $c_{n}$ defined as...

$\displaystyle c_{n} = \frac{1}{n}\ \sum_{k=1}^{n} a_{k}\ (1)$

... if $\displaystyle \lim_{n \rightarrow \infty} a_{n}=A$ then $\displaystyle \lim_{n \rightarrow \infty} c_{n}=A$. If You substitute the $a_{n}$ with $\ln a_{n}$ and L with ln L in your expression and apply Cesaro theorem, You solve the problem...Kind regards $\chi$ $\sigma$

 

[mathgirl1]

Thank you! That should get me started somewhere. Our book doesn't have the Cesaro Theorem in it, nor did he cover it so not sure I can use that method without proving it and the proof looks quite lengthy. But I guess I at least have a starting point. Confused why he would give us something like this now...

Thanks again!

If anyone has any suggestions on how to prove this without Cesaro I would be very interested in seeing that.

 

[chisigma]

Thank you! That should get me started somewhere. Our book doesn't have the Cesaro Theorem in it, nor did he cover it so not sure I can use that method without proving it and the proof looks quite lengthy. But I guess I at least have a starting point. Confused why he would give us something like this now...

Thanks again!

If anyone has any suggestions on how to prove this without Cesaro I would be very interested in seeing that.

All right mathgirl!... using logarithms we set...

$\displaystyle \lambda_{n} = \ln a_{n} = \Lambda + \gamma_{n}\ (1) $

... where $\Lambda = \ln L$ and $\displaystyle \lim_{n \rightarrow \infty} \gamma_{n}= 0$. Now we can write...

$\displaystyle w_{n} = \ln \sqrt[n]{a_{1}\ a_{2}\ ...\ a_{n}} = \Lambda + \frac{1}{n} \sum_{k=1}^{n} \gamma_{k}\ (2)$

... and from (2) You can conclude that is $\displaystyle \lim_{n \rightarrow \infty} w_{n} = \Lambda$...

Kind regards

$\chi$ $\sigma$

 

[Ackbach]

I can't say I know how to fully solve your problem, but chisigma's suggestions all look like the proof of the AM-GM inequality, which depend on the fact that the logarithm function is a strictly concave function. I'd have thought you could use AM-GM to show that the limit is less than or equal to $L$. Not sure how you could get greater than or equal to, though.

 

[mathgirl1]

Thanks so much! That helped alot!