Showing a particular recurrence is constant A sequence, $ ( a_n ) _ { n  \in \mathbb{N}} $, is constructed by selecting a value of $ a_0$, and then successively forming the following elements from the equation.
$$
a_n = 2- \frac12 a_ { n- 1}
$$
for all $ n \geq 1$. This sequence generates the power series
$$
\sum_{n=0}^{\infty}a_nx^n.
$$
Select $ a_0 = \frac43 $. Show by induction that then the sequence $(a_n)_{n \in \mathbb{N}}$ is constant.
How can this be showed with induction? What will be the basis and the inductive step?
 A: Here is a way to make use of the power series. We have
\begin{align}
F(x) & = \sum_{n=0}^{\infty} a_nx^n = a_0 + \sum_{n=1}^{\infty} a_nx^n = a_0 + \sum_{n=1}^{\infty}\left(2-\dfrac{a_{n-1}}2\right)x^n = a_0 + 2\sum_{n=1}^{\infty}x^n - \dfrac12 \sum_{n=1}^{\infty}a_{n-1}x^n\\
& = a_0 + \dfrac{2x}{1-x} - \dfrac{x}2 F(x)
\end{align}
Hence,
\begin{align}
\left(1+\dfrac{x}2\right)F(x) & = \dfrac43 + \dfrac{2x}{1-x} = \dfrac{2x+4}{3(1-x)} \implies \dfrac{(x+2) F(x)}2 = \dfrac23 \cdot \dfrac{x+2}{1-x} \end{align}
\begin{align}
F(x)
& = \dfrac43 \cdot \dfrac1{1-x} =\dfrac43 \cdot \sum_{n=0}^{\infty}x^n
\end{align}
This gives us that $a_n = 4/3$ for all $n \in \mathbb{N}$.
A: The power series is irrelavant here. Just assume $a_{n-1}=\frac 43$ and compute $a_n$.
A: This induction proof will show that each $a_n=\frac 43$ for each $n \in \mathbb{N}$, which would mean $(a_n)_{n \in \mathbb{N}}=(\frac 43)_{n \in \mathbb{N}}$ is a constant sequence.
Base step: For $n=1$, we show that $a_1=2-\frac 12 a_0 = 2-\frac 12 (\frac 43)=\frac 43$.
Inductive step: Assume for $n=k$ that we have $a_k=\frac 43$. Show for $n=k+1$ that $a_{k+1}=\frac 43$ as well. 
A: You can show by induction that $a_n = \frac{4}{3}$ for all $n=0,1,2,\dots$.
The base case is $a_0 = \frac{4}{3}$ which is given.
The inductive step is to show that $a_n = \frac{4}{3}$ assuming you have $a_{n-1} = \frac{4}{3}$. This is just a calculation:
$$a_n = 2-\frac{1}{2}\times \frac{4}{3} = \frac{6}{3}-\frac{2}{3} = \frac{4}{3}.$$
