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If $\displaystyle \lim_{n\to\infty}|a_{n+1}/a_n|=L$, then $\displaystyle\lim_{n\to\infty}|a_n|^{1/n}=L$.

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 $$ (L-\epsilon)^n\frac{|a_1|}{L-\epsilon}<|a_n|<(L+\epsilon)^n\frac{|a_1|}{L+\epsilon}.$$

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 $\lim_{n\to\infty}s_n=0$ to prove the statement.

I don't know how to use the hint. Any help?

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HINT: Let $b_n = \ln\left|\frac{a_{n+1}}{a_n}\right| - L$. Note that by hypothesis $b_n\to 0$, 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 $n\to\infty$? (There's a slight issue with $n-1$ terms instead of $n$, but you can remedy that easily.)

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  • $\begingroup$ I got it, thanks! $\endgroup$ – Cathy Jul 22 '20 at 19:43
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The key to the hint is that $$ \lim_{n\rightarrow\infty} \left(\ln\left|\frac{a_{n+1}}{a_{n}}\right|-\ln L\right) = 0. $$ Can you figure it out from here?

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    $\begingroup$ You have the fraction upside down here. To the OP: So it might help to let $b_n = \ln\left|\frac{a_{n+1}}{a_n}\right| - \ln L$ and rewrite the hint. $\endgroup$ – Ted Shifrin Jul 22 '20 at 18:46
  • $\begingroup$ Thanks, but I still don't know how to use $\displaystyle s_n=\frac{a_1+a_2+\dots+a_n}{n}$ $\endgroup$ – Cathy Jul 22 '20 at 18:53
  • $\begingroup$ @TedShifrin, thanks, but I still don't know how to solve this question after let $b_n = \ln\left|\frac{a_{n+1}}{a_n}\right| - \ln L$ $\endgroup$ – Cathy Jul 22 '20 at 19:23
  • $\begingroup$ HINT: Use the properties of $\ln$ to simplify $b_1+b_2+\dots+b_n$. (When I said rewrite the hint, I meant write it with $b_j$ in it.) $\endgroup$ – Ted Shifrin Jul 22 '20 at 19:27
  • $\begingroup$ @TedShifrin I still cannot get it, could you elaborate a little more? $\endgroup$ – Cathy Jul 22 '20 at 19:31

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