Merge pull request #75 from cren5/patch-3

Update chapter2-3.tex

Author: jesseli2002

chapters/chapter2/chapter2-3.tex

@@ -194,7 +194,7 @@ \section{The Algebraic and Order Limit Theorems}
   \item We can't use the ALT since $a_n$ is not necessarily convergent. $a_n$ being bounded gives $|a_n| \le M$ for some $M$ giving
     $$|a_nb_n| \le M|b_n| < \epsilon$$
     Which can be accomplished by letting $|b_n| < \epsilon/M$ since $(b_n) \to 0$.
-  \item No
+  \item If $a_n$ converges, then by ALT $a_n b_n$ converges. If $a_n$ diverges, then $a_n b_n$ diverges. The second conclusion comes from the fact that $b_n$ is eventually nonzero, so multiplying by the terms of a divergent sequence will result in a divergent sequence (which would not be the case for $b=0$).
   \item In (a) we showed $\lim (a_nb_n) = 0 = ab$ for $b = 0$ which proves part (iii) of the ALT.
   }
 \end{solution}