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?

Put $a_n=\frac{n+2}{n+1}b_n$. Then $b_n=b_1=\frac{1}{2}$, so $a_n=\frac{n+2}{2n+2}$ and $\lim \limits_{n\to\infty}a_n=\frac{1}{2}$.
