From 151588bcb60e7f5c711abcbdcc6c6c6a7bfd6b42 Mon Sep 17 00:00:00 2001
From: ceciliachan1979 <75919064+ceciliachan1979@users.noreply.github.com>
Date: Tue, 5 Mar 2024 06:39:52 -0800
Subject: [PATCH] Fix the same problem again

---
 Books/BabyRudin/Chapter03/ex06.cecilia.tex | 8 ++++----
 1 file changed, 4 insertions(+), 4 deletions(-)

diff --git a/Books/BabyRudin/Chapter03/ex06.cecilia.tex b/Books/BabyRudin/Chapter03/ex06.cecilia.tex
index 6c8acae..5e8dec4 100644
--- a/Books/BabyRudin/Chapter03/ex06.cecilia.tex
+++ b/Books/BabyRudin/Chapter03/ex06.cecilia.tex
@@ -49,12 +49,12 @@ \subsubsection*{Part d}
 \begin{eqnarray*}
   & & \frac{\frac{1}{|z^{n+1}| - 1}}{\frac{1}{|z^n| - 1}} \\
   &=& \frac{|z^n| - 1}{|z^{n+1}| - 1} \\
-  &<& \frac{|z^n| - 1}{|z^{n+1}|} \\
-  &=& \frac{|z^n|}{|z^{n+1}|} - \frac{1}{|z^{n+1}|} \\
-  &=& \frac{1}{|z|} - \frac{1}{|z^{n+1}|}
+  &<& \frac{|z^n| - 1}{|z^{n+1}| - |z|} \\
+  &=& \frac{1}{|z|} \\
+  &<& 1
 \end{eqnarray*}
 
-So, as $ n $ tends to $ \infty $, the sequence $ \frac{a_{n+1}}{a_n} $ is upper bounded by a sequence that tends to $ \frac{1}{|z|} < 1 $, so the sequence converges by the ratio test.
+Therefore the $ \limsup \left| \frac{\frac{1}{|z^{n+1}| - 1}}{\frac{1}{|z^n| - 1}} \right| < 1$.
 
 On the other hand, if $ |z| \ge 1 $.