Recurrence sequence limit I would like to find the limit of 
$$
\begin{cases}
a_1=\dfrac3{4} ,\, & \\
a_{n+1}=a_{n}\dfrac{n^2+2n}{n^2+2n+1}, & n \ge 1
\end{cases}
$$
I tried to use this - $\lim \limits_{n\to\infty}a_n=\lim \limits_{n\to\infty}a_{n+1}=L$, so $L=\lim \limits_{n\to\infty}a_n\dfrac{n^2+2n}{n^2+2n+1}=L\cdot1$ What does this result mean? Where did I make a mistake?

$\begin{cases}
a_1=\dfrac3{4} ,\, & \\
a_{n+1}=a_{n}\dfrac{n^2+2n}{n^2+2n+1} & n \ge 1
\end{cases}
$
$a_{n+1}
=a_{n}\left(\dfrac{n^2+2n}{n^2+2n+1}\right)
=a_{n}\left(\dfrac{n(n+2)}{(n+1)^2}\right)
$
so
$\begin{align}
a_{n}
&=a_1\prod \limits_{k=1}^{n-1} \dfrac{a_{k+1}}{a_k}\\
&=a_1\prod \limits_{k=1}^{n-1} \dfrac{k(k+2)}{(k+1)^2}\\
&=a_1 \dfrac{\prod \limits_{k=1}^{n-1}k\prod \limits_{k=1}^{n-1}(k+2)}{\prod \limits_{k=1}^{n-1}(k+1)^2}\\
&=a_1 \dfrac{\prod \limits_{k=1}^{n-1}k\prod \limits_{k=3}^{n+1}k}{\prod \limits_{k=2}^{n}k^2}\\
&=a_1 \dfrac{\prod \limits_{k=1}^{n-1}k}{\prod \limits_{k=2}^{n}k}\dfrac{\prod \limits_{k=3}^{n+1}k}{\prod \limits_{k=2}^{n}k}\\
&=a_1 \dfrac{1}{n}\dfrac{n+1}{2}\\
&=a_1 \dfrac{n+1}{2n}\\
\end{align}
$
Since $a_1 = \dfrac34$,
$a_n
=\dfrac34 \dfrac{n+1}{2n}
= \dfrac{3(n+1)}{8n}
$.
