the real sequence $\{p_n\}$ is defined as $p_1=2, p_{n+1}=( 2+p_n)/(2p_{n}+1)$ please show the procedure of this question, I am not sure how to start.
Note that if we let $f(z) = \frac{z+2}{2z +1}$ then $p_n = f^{n-1}(p_1)$. $f$ is a Moebius transform, so we can determine its powers by determining the powers of its associated matrix, $$ A_f=\begin{pmatrix} 1 &amp; 2 \\ 2 &amp; 1\end{pmatrix}.$$

You can work out the powers of this matrix in a few ways, the Jordan decomposition being the most usual, but a more direct approach, particularly useful when dealing with two by two symmetric matrices, is to write $$A_f = 3 \begin{pmatrix} 1/2 &amp; 1/2 \\ 1/2 &amp; 1/2\end{pmatrix} - \begin{pmatrix} 1/2 &amp; -1/2 \\ -1/2 &amp; 1/2\end{pmatrix} = 3R -S$$ and note that $R^2 = R$, $S^2=S$ and $RS = SR = 0$. Therefore $$A_{f^n} = A_f^n = 3^n R + (-1)^nS = \frac{1}{2}\begin{pmatrix} 3^n +(-1)^n &amp; 3^n -(-1)^n \\ 3^n -(-1)^n &amp; 3^n +(-1)^n\end{pmatrix}.$$ Therefore, $$p_n = f^{n-1}(2) = \frac{ (3^{n-1} + (-1)^{n-1})\cdot 2 + (3^{n-1} - (-1)^{n-1}) }{ (3^{n-1} - (-1)^{n-1})\cdot 2 + (3^{n-1} + (-1)^{n-1}) } = \frac{3^n + (-1)^{n-1}}{3^n - (-1)^{n-1} },$$ and the limit is easy to calculate from there.

This is almost certainly not how this question was meant to be answered.
Observe that $p_n &gt; \dfrac{1}{2}, \forall n \geq 1$. If $L$ is the limit, then $L = \dfrac{2+L}{2L+1}\to 2L^2 = 2 \to L = -1,1$. Since $L &gt; 0$, we have $L = 1$. This led us to define: $|p_{n+1} - 1| = \left|\dfrac{2+p_n}{2p_n+1} - 1\right| = \left|\dfrac{2+p_n - 2p_n - 1}{2p_n+1}\right|= \left|\dfrac{1-p_n}{2p_n+1}\right|&lt;\dfrac{|1-p_n|}{2}&lt; ...&lt; \dfrac{|1-p_1|}{2^n}= \dfrac{1}{2^n} \to 0$. Thus $p_{n+1} \to 1$ as claimed.