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--- math104-f21:hw4 [2021/09/16 10:45]
pzhou created
+++ math104-f21:hw4 [2022/01/11 08:36] (current)
pzhou ↷ Page moved from math104:hw4 to math104-f21:hw4
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 . Let  $(a_n)$  be a bounded sequence in  $\R$ , and  $A = \lim sup(a_n)$ , show that for any  $\epsilon > 0$ , the set  $\{ n \in \N \mid A - \epsilon \leq a_n \leq A+\epsilon \}$  is infinite.
+====== Solution ======
+. Suffice to show that, for any $\epsilon>0$, there is an  $N$ , such that for all  $n>N$ , we have
+ |a_n - x| \leq \epsilon,
+which is
+ $x - \epsilon \leq a_n \leq x + \epsilon.$
+By Cauchy condition for  $a_n$ , take  $\epsilon_1 = \epsilon$ , there exists  $N_1 > 0$ , such that for all  $n,m> N_1$ , we have  $|a_n - a_m| < \epsilon_1$ , in particular, we have
+ $a_m - \epsilon \leq a_n \leq a_m + \epsilon.$
+Consider the sequence  $(a_m)_{m = N_1}^\infty$ , then we have its formal limit satisfies, for all  $n > N_1$ ,
+ $x - \epsilon = (LIM a_m) - \epsilon \leq a_n \leq (LIM a_m) + \epsilon = x + \epsilon.$
+Finally, take  $N = N_1$ .
+. Take any  $N > 2|a|$ , then for any  $n > N$ , we have
+ $|a_{n} / a_{n-1}| = |a| / n < 1/2$
+Hence $|a_{n}| \leq |a_{N}| (1/2)^{n-N}$, take limit  $n \to \infty$  shows the result.
+. Since  $A$  is bounded above, we know the  $x=sup(A)$  exists. Suppose  $x^2 < 3$ , then by the same argument as in Prop 5.5.12, we have a small enough $\epsilon>0$, such that  $(x+\epsilon)^2 < 3$ , then let  $q$  be a rational number, such that  $x < q < x+\epsilon$ , we have  $q^2 < 3$ , hence  $q \in A$ . This contradicts with with  $x$  being the  $sup$  of  $A$ . Suppose  $x^2 > 3$ , then we can find small enough $\epsilon>0$, such that  $(x-\epsilon)^2 > 3$ , in particular  $x-\epsilon$  is also an upperbound of  $A$ , hence contradict with  $x$  being the upper bound, hence  $x^2 = 3$ .
+. Since  $s_{n+1}^2 = 1 + s_n$ , and  $(s_n)$  converges, hence
+ $\lim s_{n+1}^2 = \lim(1 + s_n)$
+which gives
+ $c^2 = 1 + c$
+solving this gives $c = (\pm \sqrt{5}+1)/2 . Since  $s_n$  are all positive by induction and, we see $c = ( \sqrt{5}+1)/2$
+. This is covered in last Friday's lecture, where we proved that  $L=\limsup(a_n)$  is a limit point of  $a_n$ . Here we give the proof again.
+We prove by contradiction, suppose there is an $\epsilon>0$, such that there are only finitely many terms in  $(a_n)$  contained in $[L-\epsilon, L+\epsilon] $. Then there exists an$ N_1 $, such that for all$ n>N_1 $,$ |a_n - L| > \epsilon $. By property of$ \limsup $, we know there also exists$ N_2 $, such that for all$ n > N_2 $,$ a_n < L + \epsilon $. Then, for$ n > N= \max(N_1, N_2)$, we have
+ |a_n - L| > \epsilon,  \quad a_n < L + \epsilon \Rightarrow a_n < L - \epsilon.
+Taking  $\limsup$  on the left, we get  L \leq L - \epsilon$, which is a contradiction.

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