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?
 A: You didn't make any mistake with the limits, you get $L=L$, though uninformative, it is true. 
You can prove by induction that $\forall n\in \Bbb N\left(a_n=\dfrac 3 8\dfrac{n+1}{n}\right)$.
Inductive step: Let $n\in \Bbb N$ be such that $a_n=\dfrac 3 8\dfrac{n+1}{n}$. Then the following holds:
$$\begin{align} a_{n+1}=a_n\dfrac{n^2+2n}{n^2+2n+1}&=\dfrac 3 8\dfrac{n+1}{n}\dfrac{n^2+2n}{n^2+2n+1}\\
&=\dfrac 3 8\dfrac{n+1}{n}\dfrac{n(n+2)}{(n+1)^ 2}\\
&=\dfrac 3 8\dfrac{n+2}{n+1}.\end{align}$$
A: 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 
$$
\lim_{n\to\infty}a_n=\frac{3}{8}.
$$
A: $\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}
$.
A: 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 $\lim \limits_{n\to\infty}a_n=\frac{1}{2}$.
