<s>[INST] How to find limit of sequence $a_1=1$, $a_{n+1}=\left( 1-\frac{1}{(n+1)^2} \right)a_n$ for $n\geq 1$? I need some help finding this limit.

Define a sequence by $a_1=1$ and
  $a_{n+1}=\left(1-\frac{1}{(n+1)^2}\right)a_n$ for $n\geq1$. Show that the limit exists and find the limit.

I've shown that the limit $L$ exists by showing it's decreasing and bounded below by $0$, but I'm not sure how to actually find it. I couldn't get a general formula for $a_n$. Also, the trick where I write $L=\lim_{n\to\infty}a_{n+1}$ and substitute the formula for $a_{n+1}$ doesn't seem to work here.
I calculated the first few terms and it seems like $L$ is $1/2$, but otherwise I'm stuck here and not sure how to proceed.
 [/INST] $1-\frac 1{k^2} = \frac{(k-1)(k+1)}{k^2}$, so the product is
$$
\frac{(2-1)(2+1)}{2\cdot 2}\cdot\frac{(3-1)(3+1)}{3\cdot 3}\cdot\frac{(4-1)(4+1)}{4\cdot 4}\cdot\frac{(5-1)(5+1)}{5\cdot 5}\cdot\dots
$$
i.e.,
$$
\frac{3}{2\cdot 2}\cdot\frac{2\cdot 4}{3\cdot 3}\cdot\frac{3\cdot 5}{4\cdot 4}\cdot\frac{4\cdot 6}{5\cdot 5}\cdot\dots
$$
Do you see something?
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