# Let ${a_{n + 1}} = \frac{{{a_n} - 1}}{{{a_n} + 1}}.$ How can we found general term of this sequence? [closed]

Let

• $\left\{ {{a_n}} \right\}$ be a sequence.

• $\alpha \in \mathbb{R} .$

• $a_1=\alpha$

• ${a_{n + 1}} = \frac{{{a_n} - 1}}{{{a_n} + 1}}.$

How can we found general term of this sequence?

• Have you tried starting from $n=0$, then $n=1$, etc, and writing out the expressions you obtain? – user137731 Oct 25 '16 at 18:56
• Is the second point to mean $\;a_0=\alpha\;$ ? (or $\;a_1\;$, it never matters) – DonAntonio Oct 25 '16 at 18:57
• @DonAntonio - I'm sorry. I edited this post. – Under sky Oct 25 '16 at 19:00
• Hint: compute $a_2, a_3, a_4, a_5$ by hand, what is $a_5$? – achille hui Oct 25 '16 at 19:08
• I'm voting to close this question as off-topic because the person who asked is refusing to do anything at all. Just refusing. People are giving good advice, including suggestions for how to investigate and actually learn something through the experience – Will Jagy Oct 25 '16 at 19:25

Assume that $a_n=\frac{p_n}{q_n}$ is represented by the vector $v_n=(p_n,q_n)^T\in\mathbb{R}^2$. Then $$a_{n+1} = \frac{p_n-q_n}{p_n+q_n}$$ is represented by the vector $$v_{n+1}=\begin{pmatrix}p_{n+1}\\ q_{n+1}\end{pmatrix} = \begin{pmatrix}1 & -1 \\ 1 & 1\end{pmatrix}v_n$$ so we may solve the problem by finding an explicit expression for $M^n$ where $M=\begin{pmatrix}1 & -1 \\ 1 & 1\end{pmatrix}$.
Since the eigenvalues of $M$ are $1+i$ and $1-i$, by diagonalizing $M$ we get that $$p_n = A(1+i)^n + B(1-i)^n,\qquad q_n = C(1+i)^n+D(1-i)^n$$ for a set of constants $A,B,C,D$ that depend on $p_0,q_0,p_1,q_1$. By imposing $p_1=\alpha,q_1=1$, $p_2=\alpha-1$, $q_2=\alpha+1$ it is easy to find $A,B,C,D$, then the general expression for $a_n$: $$a_n = \color{red}{\frac{(1-i\alpha)+(\alpha-i)(-i)^n}{(1+i\alpha)(-i)^n-(\alpha+i)}}.$$ We may notice that since $M^4=-4I$, $\;\color{red}{a_n=a_{n+4}}$, so the whole sequence is made by the terms $\alpha,\frac{\alpha-1}{\alpha+1},-\frac{1}{\alpha}$ and $\frac{1+\alpha}{1-\alpha}$ only.