Finding $\lim\limits_{n \to \infty} \sqrt[n^2]{a_n}$ given that $\lim\limits_{n \to \infty} \frac{a_n a_{n+2}}{(a_{n+1})^2}=L^2$ 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.
