A sequence $\lbrace a_{n}\rbrace_{n\geq 0}$ is constructed by choosing a value of $a_{n}$, and then the following elements are determined from the equation $a_{n}=2-\frac{1}{2}a_{n-1}$ for $n\geq 1$. Let the power series be given $\sum_{n\geq 0}a_{n}x^{n}$.
a) Choose $a_{0}=4/3$. Show by induction that $a_{n}$ is constant, and determine the interval of convergence and the sum function of the power series.
My answer: First I see that $a_{1}=4/3$ so there must be $a_{n}=4/3$ for all $n\geq 1$. For $n=1$ the statement is clearly true. Assume that the statement is true for $n=m$. For $n=m+1$ we have $a_{m+1}=2-(1/2)a_{m}=2-(1/2)a_{n}=a_{n}$, so the statement for $n=m+1$ is true. We have shown that it is constant.
We have the power series $\sum_{n\geq 0}(4/3)x^{n}$ which is a geometric series that converges if $x\in (-1,1)$ with sum $(4/3)(1-x)^{-1}$.
b) Choose $a_{0}=2/3$. Show by induction that it holds $2/3\leq a_{n}\leq 5/3$ for all $n$, and use it to determine the radius of convergence of the power series.
My answer: First, we see that $$\frac{2}{3}\leq a_{n}\leq \frac{5}{3}\iff \frac{2}{3}\leq 2-\frac{1}{2}a_{n-1}\leq \frac{5}{3}\iff \frac{2}{3}\leq a_{n-1}\leq \frac{8}{3}$$ for all $n\geq 1$. Let $a_{0}=\frac{2}{3}$. We will start showing that $\frac{2}{3}\leq a_{n}$ by induction. The statement is true for $n=1$, since $\frac{2}{3}\leq a_{1}=2-\frac{1}{2}a_{0}$. Assuming $n=m$ to be true and for $n=m+1$, we have $$\frac{2}{3}\leq a_{m+1}=2-\frac{1}{2}a_{m}=2-\frac{1}{2}a_{n}=2-\frac{1}{2}\left [ 2-\frac{1}{2}a_{n-1} \right ]=1+\frac{1}{4}a_{n-1}\leq 1+\frac{1}{4}\frac{8}{3}=\frac{5}{3}$$ which is true. We will now show that $a_{n}\leq \frac{5}{3}$. The statement $n=1$ is clear. Assuming $n=m$ to be true and for $n=m+1$ we have $$-\frac{5}{3}\geq -a_{m+1}=-\left [ 2-\frac{1}{2}a_{m} \right ]=\dots=-\left [ 1+\frac{1}{4}a_{n-1} \right ]\geq -\frac{5}{3}.$$ Since it's true for $n=m+1$, we conclude that the hypothesis is true for all $n\geq 1$ by induction.
Since $\frac{2/3}{5/3}\leq \left | \frac{a_{n}}{a_{n+1}} \right |\leq \frac{5/3}{2/3}$, so the convergence of radius is $r=\lim_{n\to\infty} \left | \frac{a_{n}}{a_{n+1}} \right |= \frac{5}{2}$.
How are my answers so far?
Let $A(x) =\sum_{n\geq 0}a_{n}x^{n} $. Since $a_{n}=2-\frac{1}{2}a_{n-1}$,
$\begin{array}\\ A(x) &=\sum_{n\geq 0}a_{n}x^{n}\\ &=a_0+\sum_{n\geq 1}a_{n}x^{n}\\ &=a_0+\sum_{n\geq 1}(2-\frac{1}{2}a_{n-1})x^{n}\\ &=a_0+\sum_{n\geq 1}2x^n-\frac{1}{2}\sum_{n\geq 1}a_{n-1}x^{n}\\ &=a_0+\frac{2x}{1-x}-\frac{x}{2}\sum_{n\geq 1}a_{n-1}x^{n-1}\\ &=\frac{2x+a_0-xa_0}{1-x}-\frac{x}{2}\sum_{n\geq 0}a_{n}x^{n}\\ &=\frac{x(2-a_0)+a_0}{1-x}-\frac{x}{2}A(x)\\ \end{array} $
so $A(x)(1+\frac{x}{2}) =\frac{x(2-a_0)+a_0}{1-x} $ or $A(x) =\frac{x(2-a_0)+a_0}{1-x}\frac{2}{2+x} $.
In general, the radius of convergence is $2$ (from the $x+2$). However, if $a_0 = 2(2-a_0)$, so $a_0 = \frac43$, $A(x) =\frac{2(\frac43+\frac23 x)}{(1-x)(2+x)} =\frac{\frac43(2+ x)}{(1-x)(2+x)} =\frac{4/3}{1-x} $, the radius of convergence is $1$.
Note that we can never cancel out the $1-x$ in the denominator, because this would require that $a_0 = -(2-a_0) =-2+a_0$, which is impossible.