This is the problem I saw in another question - which was later closed and deleted (the close reason was missing context). Still, the problem seemed interesting to me - at least in the sense that the solution is not immediately obvious.

The problem:

Suppose that $(a_n)$ is a real sequence such that $a_n>0$ for $n>0$ and $$\lim\limits_{n\to\infty} \frac{a_na_{n+2}}{a_{n+1}^2}=L^2.$$ Suppose further that $L\ge0$ Find $\lim\limits_{n\to\infty} \sqrt[n^2]{a_n}$. (In terms of $L$.)

The original problem in the linked question was given with $L=2$. However, I do not think that the problem should change that much for any value of $L$. (Maybe one could be suspicious about $L=0$?)

It is also clear - at least for $L>0$ - that if it is possible to express the second limit using $L$, then the limit should be equal to $L$. It suffices to notice that for $a_n = L^{n^2}$ we have $$\frac{a_na_{n+2}}{a_{n+1}^2}= \frac{L^{n^2+(n+2)^2}}{L^{2(n+1)^2}} = L^{(2n^2+4n+4)-2(n^2+2n+2)} = L^2. $$

So we know what is the candidate for the result, we still need to prove whether this is actually true.

I have posted my attempt as answer. But I will be glad to learn about other solutions. (And of course, also corrections to my approach, if I made some mistakes.)


For the sake of simplicity, let us assume that $L>0$. And later we will check what has to be modified in the case $L=0$.

Let us denote $b_n = \frac{\log a_n}{\log L}$. (I.e., the logarithm at the base $L$.) Then the limit $$\lim\limits_{n\to\infty} \frac{a_na_{n+2}}{a_{n+1}^2}=L^2$$ changes to $$\lim\limits_{n\to\infty} (b_{n+2}-2b_{n+1}+b_n) = 2.$$ And if we put $c_n=b_{n+1}-b_n$ then we have $$\lim\limits_{n\to\infty} (c_{n+1}-c_n)=2.$$ Applying Stolz-Cesaro theorem1 twice we get $$\lim\limits_{n\to\infty} \frac{b_{n+1}-b_n}n = \lim\limits_{n\to\infty} \frac{c_n}n =2$$ and $$\lim\limits_{n\to\infty} \frac{b_n}{\frac{n(n-1)}2}=2.$$

Therefore we get $$\lim\limits_{n\to\infty} \frac{b_n}{n(n-1)}=1 \qquad\text{and}\qquad \lim\limits_{n\to\infty} \frac{b_n}{n^2} = 1.$$ So we have $$\lim\limits_{n\to\infty} \frac{\log{a_n}}{n^2\log L} = \frac{\log{\sqrt[n^2]{a_n}}}{\log L}=1$$ i.e. $$\lim\limits_{n\to\infty} \log \sqrt[n^2]{a_n}=\log L \qquad\text{and}\qquad \lim\limits_{n\to\infty} \sqrt[n^2]{a_n}=L.$$

What happens if $L=0$? Already the first step does make sense, since we cannot take logarithm of $L^2=0$.

Let us now take $b_n=\log a_n$. (To avoid $\log L$ in denominator-- which was there just to simplify the numbers we work with, so that we do not have to write $\log L$ every time.)

In the first step we have just used that $x_n\to L$ implies $\log x_n\to\log L$. If we have $x_n\to0^+$, we still have $\log x_n\to -\infty$. So for $b_n=\log a_n$ we $\lim\limits_{n\to\infty} (b_{n+2}-2b_{n+1}+b_n) = -\infty$.

Stolz-Cesaro theorem is true also for $\pm\infty$. Which means that the following steps are correct if we replace $2$ by $-\infty$.

Finally we arrive at $\lim\limits_{n\to\infty} \log \sqrt[n^2]{a_n}=-\infty$, which gives us $$\lim\limits_{n\to\infty} \sqrt[n^2]{a_n}=0$$ simply by using that $\log x_n\to-\infty$ implies $x_n\to0^+$.

Which means that the same result is true for $L=0$.

1See Wikipedia: Stolz–Cesaro theorem. Some proofs of this result can be found in this question.

  • $\begingroup$ Just one doubt. The question was asked 3 Minutes ago. And you posted such a long answer 3 minutes ago. How is this even possible? $\endgroup$ – user567182 Jun 28 '18 at 8:38
  • $\begingroup$ @user567182 It is possible to post a question and an answer at the same time. This feature was mentioned on meta around the time when it was introduced: Recently rolled out SE Encyclopedia feature. In which situations it is suitable is another question - but it is definitely encouraged that the poster shows their own attempts when posting the question, posting self-answer might be a way to do this. If you are interested in other past discussions about this, you might have a look at posts tagged (self-answer) on meta. $\endgroup$ – Martin Sleziak Jun 28 '18 at 8:42
  • $\begingroup$ Perfect answer. +1 I had mentioned the same approach in comments to the original (now closed deleted question). $\endgroup$ – Paramanand Singh Jun 28 '18 at 14:32

Here is an answer that I posted to the question mentioned in this question, but deleted almost immediately due to the lack of context in that question.

If $$ \lim_{n\to\infty}\frac{a_n\,a_{n+2}}{a_{n+1}^2}=L^2\tag1 $$ then for any $\epsilon\gt0$, there is an $N_\epsilon$ so that for $n\ge N_\epsilon$, $$ \left|\,\log\left(\frac{a_n\,a_{n+2}}{a_{n+1}^2}\right)-2\log(L)\,\right|\le\epsilon\tag2 $$ Note that $$ \begin{align} \frac1{n^2}\sum_{k=0}^{n-1}\sum_{j=0}^{k-1}\log\left(\frac{a_j\,a_{j+2}}{a_{j+1}^2}\right) &=\frac1{n^2}\sum_{k=0}^{n-1}\log\left(\frac{a_0}{a_k}\frac{a_{k+1}}{a_1}\right)\\ &=\frac1n\log\left(\frac{a_0}{a_1}\right)+\frac1{n^2}\log\left(\frac{a_n}{a_0}\right)\tag3 \end{align} $$ Inspired by the following diagram

enter image description here

We can break down the sum on the left side of $(3)$ as follows $$ \begin{align} \frac1{n^2}\sum_{k=0}^{n-1}\sum_{j=0}^{k-1}\log\left(\frac{a_j\,a_{j+2}}{a_{j+1}^2}\right) &=\frac1{n^2}\sum_{k=N_\epsilon}^{n-1}\sum_{j=N_\epsilon}^{k-1}\overbrace{\log\left(\frac{a_j\,a_{j+2}}{a_{j+1}^2}\right)}^\text{within $\epsilon$ of $2\log(L)$}\\ &\color{#C00}{+\frac{n-N_\epsilon}{n^2}\sum_{j=0}^{N_\epsilon-1}\log\left(\frac{a_j\,a_{j+2}}{a_{j+1}^2}\right)+\frac1{n^2}\sum_{k=0}^{N_\epsilon-1}\sum_{j=0}^{k-1}\log\left(\frac{a_j\,a_{j+2}}{a_{j+1}^2}\right)}\tag4 \end{align} $$ For any $\epsilon\gt0$, the red terms vanish as $n\to\infty$. The remaining term on the right hand side is $\frac1{n^2}$ times a sum of $\frac{(n-N_\epsilon)^2+(n-N_\epsilon)}2$ terms, all within $\epsilon$ of $2\log(L)$.

Therefore, $$ \left|\,\lim_{n\to\infty}\frac1{n^2}\sum_{k=0}^{n-1}\sum_{j=0}^{k-1}\log\left(\frac{a_n\,a_{n+2}}{a_{n+1}^2}\right)-\log(L)\,\right|\le\frac\epsilon2\tag5 $$ Since $(5)$ is true for any $\epsilon\gt0$, we have $$ \lim_{n\to\infty}\frac1{n^2}\sum_{k=0}^{n-1}\sum_{j=0}^{k-1}\log\left(\frac{a_n\,a_{n+2}}{a_{n+1}^2}\right)=\log(L)\tag6 $$ Taking the limit of the right hand side of $(3)$ as $n\to\infty$, we get $$ \lim_{n\to\infty}\frac{\log(a_n)}{n^2}=\log(L)\tag7 $$ which implies $$ \lim_{n\to\infty}\sqrt[n^2]{a_n}=L\tag8 $$

  • $\begingroup$ It's great that you posted your answer here. By the time I saw your answer (to the old deleted question) it was in deleted state. +1 $\endgroup$ – Paramanand Singh Jun 29 '18 at 8:47
  • $\begingroup$ Not surprising, since it was only there for 18 seconds. $\endgroup$ – robjohn Jun 29 '18 at 11:50

This is not another solution, but is the essence of your solution worked out by me individually.

If $L>0$,

By two applications of Stolz-Cesaro theorem,

\begin{align*} \lim_{n\to\infty}{{\ln a_n}\over{n^2}} &=\lim_{n\to\infty}{\ln{\frac{a_{n+1}}{a_n}}\over{2n+1}}\\ &=\frac{1}{2}\lim_{n\to\infty}{\ln{\frac{a_{n+1}}{a_n}}\over{n}}\\ &=\frac{1}{2}\lim_{n\to\infty}{\ln{\frac{a_{n+2}}{a_{n+1}}-\ln{\frac{a_{n+1}}{a_n}}}\over1}\\ &=\frac{1}{2}\lim_{n\to\infty}\ln\Bigg(\frac{\big(\frac{a_{n+2}}{a_{n+1}}\big)}{\big(\frac{a_{n+1}}{a_{n}}\big)}\Bigg)\\ &=\frac{1}{2}\lim_{n\to\infty}\ln\frac{a_na_{n+2}}{{a_{n+1}}^2}\\ &=\frac{1}{2}\ln(L^2)\\ &=\ln L\\ \therefore\lim_{n\to\infty}\sqrt[n^2]{a_n}&=L \end{align*}

If $L=0$, \begin{align*} \lim_{n\to\infty}{{\ln a_n}\over{n^2}} &=\lim_{n\to\infty}{\ln{\frac{a_{n+1}}{a_n}}\over{2n+1}}\\ &=\lim_{n\to\infty}{\ln{\frac{a_{n+1}}{a_n}}\over{n}}\\ &=\lim_{n\to\infty}{\ln{\frac{a_{n+2}}{a_{n+1}}-\ln{\frac{a_{n+1}}{a_n}}}\over1}\\ &=\lim_{n\to\infty}\ln\Bigg(\frac{\big(\frac{a_{n+2}}{a_{n+1}}\big)}{\big(\frac{a_{n+1}}{a_{n}}\big)}\Bigg)\\ &=\lim_{n\to\infty}\ln\frac{a_na_{n+2}}{{a_{n+1}}^2}\\ &=-\infty\\ \therefore\lim_{n\to\infty}\sqrt[n^2]{a_n}&=0 \end{align*}


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