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?

It's easy to see that $a_{n+1}=a_n\left[1-\frac{1}{(n+1)^2}\right]$ and observing that
$$
\begin{align}
a_2&=a_1\left(1-\tfrac{1}{2^2}\right)\\
a_3&=a_2\left(1-\tfrac{1}{3^2}\right)=a_1\left(1-\tfrac{1}{2^2}\right)\left(1-\tfrac{1}{3^2}\right)\\
&\vdots\\
a_n&=a_{n-1}\left(1-\tfrac{1}{n^2}\right)=a_1\left(1-\tfrac{1}{2^2}\right)\left(1-\tfrac{1}{3^2}\right)\cdots\left(1-\tfrac{1}{n^2}\right)=a_1\prod_{k=2}^n\left(1-\tfrac{1}{k^2}\right)
\end{align}
$$
that is 
$$
a_n=\tfrac{3}{4}\prod_{k=2}^n\left(1-\tfrac{1}{k^2}\right).
$$
It's easy to see that
$$
\prod_{k=2}^n\left(1-\tfrac{1}{k^2}\right)=\tfrac{(n+1)}{2n}\to\frac{1}{2}\qquad\text{for }n\to\infty
$$
and finally 
$$
\lim_{n\to\infty}a_n=\frac{3}{8}.
$$
