Recurrence sequence limit

Question

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?

Answer 1

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: $$\begin{align} a_{n+1}=a_n\dfrac{n^2+2n}{n^2+2n+1}&=\dfrac 3 8\dfrac{n+1}{n}\dfrac{n^2+2n}{n^2+2n+1}\\ &=\dfrac 3 8\dfrac{n+1}{n}\dfrac{n(n+2)}{(n+1)^ 2}\\ &=\dfrac 3 8\dfrac{n+2}{n+1}.\end{align}$$

Answer 2

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 $\lim \limits_{n\to\infty}a_n=\frac{1}{2}$.

Answer 3

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 $$ \lim_{n\to\infty}a_n=\frac{3}{8}. $$

Answer 4

$\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} $.