If $lim_{ntoinfty}|a_{n+1}/a_n|=L$, then $lim_{ntoinfty}|a_n|^{1/n}=L$

Question

If $displaystyle lim_{ntoinfty}|a_{n+1}/a_n|=L$, then $displaystylelim_{ntoinfty}|a_n|^{1/n}=L$.

There're plenty of proofs available on the internet or books, such as the following one.
There exists an $N$, such that whenever $n>N$,
$$
L-epsilon<|a_n/a_{n-1}|<L+epsilon,
$$
then
$$
(L-epsilon)^nfrac{|a_1|}{L-epsilon}<|a_n|<(L+epsilon)^nfrac{|a_1|}{L+epsilon}.$$
But I'm doing a problem that specifically says that taking $log$ and using if ${a_n}$ is a sequence converging to $0$, and $displaystyle s_n=frac{a_1+a_2+dots+a_n}{n}$, then $lim_{ntoinfty}s_n=0$ to prove the statement.
I don't know how to use the hint. Any help?

Ted Shifrin · Answer

HINT: Let $b_n = lnleft|frac{a_{n+1}}{a_n}right| - L$. Note that by hypothesis $b_nto 0$, so the hint applies to that sequence.
Then
$$b_1+dots+b_{n-1} = ln|a_{n}| - ln |a_1| - (n-1)L,$$
so
$$frac{b_1+dots+b_{n-1}}n = frac{ln|a_n|}n - frac{ln|a_1|}n - frac{n-1}nL.$$
What is the limit as $ntoinfty$? (There's a slight issue with $n-1$ terms instead of $n$, but you can remedy that easily.)

eyeballfrog · Answer

The key to the hint is that
$$
lim_{nrightarrowinfty} left(lnleft|frac{a_{n+1}}{a_{n}}right|-ln Lright) = 0.
$$
Can you figure it out from here?

If $lim_{ntoinfty}|a_{n+1}/a_n|=L$, then $lim_{ntoinfty}|a_n|^{1/n}=L$

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