TITLE: Define a sequence by $a_1 = 1, a_2 = 1/2$, and $a_{n+2} = a_{n+1} - a_na_{n+1}/2$ for $n$ a positive integer.
QUESTION [7 upvotes]: Define a sequence by $a_1 = 1, a_2 = 1/2$, and $$a_{n+2} = a_{n+1} - a_na_{n+1}/2$$ for $n$ a positive integer. Find $$\lim_{n\to\infty}na_n$$ if it exists.

Well, we can deduce that $\lim a_n=0$ by checking $(a_n)$ is decreasing and bounded. But finding $\lim na_n$ is another story.

REPLY [2 votes]: $\lim a_n = 0$ and $a_n$ decreasing $\Longrightarrow b_n=\frac{1}{a_n}$ increasing and unbounded.
For the Cesàro-Stolz Theorem, if $\lim_{n\to\infty}\dfrac{(n+1)-n}{\frac{1}{a_{n+1}}-\frac{1}{a_{n}}}=L$ (*), so $\lim \dfrac{n}{\frac{1}{a_{n}}}=L$. So, we have to proof (*).
Notice that $\lim_{n\to\infty}\dfrac{(n+1)-n}{\frac{1}{a_{n+1}}-\frac{1}{a_{n}}}=\lim 2 \frac{a_{n+1}}{a_{n-1}}$. But $\lim \frac{a_{n+1}}{a_{n}}=1 \Rightarrow \lim \frac{a_{n+1}}{a_{n-1}}=1$ because $\lim a_n =0$. So, $L=2$.