I am self-studying Real Analysis from the text Understanding Analysis by Stephen Abbott. I would like for someone to 1) verify my proof for parts (a) and (b) of the below exercise problem is rigorous and correct. Do you have a hint/clue for part (c) (without giving away or revealing the entire solution)?
[Abbott 6.5.7] Let $\displaystyle \sum a_{n} x^{n}$ be a power series with $\displaystyle a_{n} \neq 0$, and assume that
\begin{equation*} L=\lim _{n\rightarrow \infty }\left| \frac{a_{n+1}}{a_{n}}\right| \end{equation*}
exists.
(a) Show that if $\displaystyle L\neq 0$, then the series converges for all $\displaystyle x$ in $\displaystyle ( -1/L,1/L)$. (The advice in exercise 2.7.9 may be helpful).
Proof.
Fix $\displaystyle x_{0} \in ( -1/L,1/L)$ be an arbitrary point.
The absolute value ratio of the successive terms of the power series at the point $\displaystyle x_{0}$ is given by:
\begin{equation*} \left| \frac{a_{n+1} x_{0}^{n+1}}{a_{n} x_{0}^{n}}\right| =\left| \frac{a_{n+1}}{a_{n}}\right| \cdot |x_{0} | \end{equation*} Passing to the limits, as $\displaystyle n\rightarrow \infty $, we have:
\begin{equation*} \lim _{n\rightarrow \infty }\left| \frac{a_{n+1}}{a_{n}}\right| \cdot |x_{0} |=L\cdot |x_{0} |< 1 \end{equation*} If $\displaystyle L\ \neq 0$, then $\displaystyle 0< L\cdot |x_{0} |=r< 1$.
By the ratio test for the convergence of an infinite series of real numbers, $\displaystyle \sum _{n=1}^{\infty } a_{n} x_{0}^{n}$ converges. Since, $\displaystyle x_{0}$ was arbitrary, this holds true for all $\displaystyle x\in ( -1/L,1/L)$. Consequently, $\displaystyle \sum _{n=1}^{\infty } a_{n} x^{n}$ converges.
(b) Show that if $\displaystyle L=0$, then the series converges for all $\displaystyle x\in \mathbf{R}$.
Proof.
We are given that $\displaystyle L=0$. So, the $\displaystyle \lim \left| \frac{a_{n+1}}{a_{n}}\right| =0$. By definition of limits, this implies:
\begin{equation*} ( \forall \epsilon >0)( \exists N\in \mathbf{N})( \forall n\geq N)\left(\left| \frac{a_{n+1}}{a_{n}}\right| < \epsilon \right) \end{equation*} Fix $\displaystyle x_{0} \in \mathbf{R}$. A power series always converges at $\displaystyle x=0$, so assume that $\displaystyle x_{0} \neq 0$.
Pick $\displaystyle \epsilon =\frac{1}{2|x_{0} |}$ and let $\displaystyle r=\frac{1}{2}$. Then, there exists $\displaystyle N\in \mathbf{N}$, such that for all $\displaystyle n\geq N$, we have: \begin{equation*} \left| \frac{a_{n+1}}{a_{n}}\right| < \frac{1}{2|x_{0} |} \end{equation*} which is
\begin{equation*} \left| \frac{a_{n+1}}{a_{n}}\right| |x_{0} |< \frac{1}{2} =r \end{equation*} Thus,
\begin{equation*} |a_{n+1} x_{0}^{n+1} |< |a_{n} x_{0}^{n} |r \end{equation*} for all $\displaystyle n\geq N$.
Consequently,
\begin{equation*} \begin{array}{ c l } \sum _{n=N}^{\infty } |a_{n} x_{0}^{n} | & =|a_{N} x_{0}^{N} |\left( 1+r+r^{2} +\dotsc \right)\\ & =\frac{|a_{N} ||x_{0}^{N} |}{1-r} \quad \left\{\text{Geometric series}\right\} \end{array} \end{equation*} Hence, $\displaystyle \sum _{n=1}^{\infty } a_{n} x_{0}^{n}$ is converges absolutely. By the absolutely convergence test, $\displaystyle \sum _{n=1}^{\infty } a_{n} x_{0}^{n}$ is a convergent series.
Since $\displaystyle x_{0}$ was arbitrary, this must be true for all $\displaystyle x\in \mathbf{R}$. Consequently, $\displaystyle \sum _{n=1}^{\infty } a_{n} x^{n}$ converges for all $\displaystyle x\in \mathbf{R}$.
(c) Show that (a) and (b) continue to hold if $\displaystyle L$ is replaced by the limit
\begin{equation*} L'=\lim s_{n} \quad \text{where} \ \quad s_{n} =\sup \left\{\left| \frac{a_{k+1}}{a_{k}}\right| :k\geq n\right\} \end{equation*}
Proof.
Fix $\displaystyle x_{0} \in \left( -\frac{1}{L'} ,\frac{1}{L'}\right)$.
Define
\begin{equation*} s_{n} =\sup \left\{\left| \frac{a_{k+1} x_{0}^{k+1}}{a_{k} x_{0}^{k}}\right| :k\geq n\right\} =\sup \left\{\left| \frac{a_{k+1}}{a_{k}}\right| |x_{0} |:k\geq n\right\} =|x_{0} |\cdot \sup \left\{\left| \frac{a_{k+1}}{a_{k}}\right| :k\geq n\right\} \end{equation*} and \begin{equation*} r_{n} =\left| \frac{a_{n+1}}{a_{n}}\right| |x_{0} | \end{equation*} We know that:
\begin{equation*} r_{n} \leq s_{n} \end{equation*}
$s_n \geq |\frac {a_{n+1}} {a_n}|$ so $L'\geq L$. This implies that $(-\frac 1 {L'}, \frac 1 {L'}) \subseteq (-\frac 1 {L}, \frac 1 L)$. This takes care of a) with $L$ repaced by $L'$.
If $L'=0$ then $L=0$ so b) is also taken care of.