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--- math104-f21:hw2-sol [2021/09/09 23:07]
pzhou created
+++ math104-f21:hw2-sol [2026/02/21 14:41] (current)
@@ Line 33: / Line 33: @@
 For integer $n \geq 1$, we have
 $$ a_{n+1} - a_n = (3 + \frac{1}{a_n}) - (3 + \frac{1}{a_{n-1}}) = \frac{a_{n-1} - a_n}{a_n a_{n-1}} $$
+Note that, for any $a_0 > 0$, we have $a_1 > 3$ and all subsequence $a_n > 3$, hence for $n \geq 2$, we have
+$$ |a_{n+1} - a_n| \leq \frac{|a_{n-1} - a_n|}{9}. $$
+Throw away the first term, we have $(a_n)_{n=1}^\infty$ satisfies the condition of problem 1, with $r=1/9$, hence it is a Cauchy sequence.
+=== 3. ===
+We need to show that, given any $\epsilon > 0$, there is an $N>0$, such that for all $n > N$, $|a_n - b_n| \leq \epsilon$. Since $(a_n)$ is Cauchy, for the given $\epsilon >0$, there exists an $N_0>0$, such that for any $n, m> N_0$, $ |a_n - a_m| \leq \epsilon$. In particular, if we take $m=2n$ and set $N=N_0$, then we get the desired result.
+=== 4. ===
+Since Cauchy sequences are bounded, there exists $M>0$, such that $|a_n| < M$ and $|b_n| < M$ for all $n$. Suppose $\epsilon > 0$ is given, we need to show that there exists $N>0$, such that for all $n,m>N$, $|a_n b_n - a_m b_m| \leq \epsilon $. We let $\epsilon_1 = \epsilon_2 = \epsilon / (2M)$, then since $(a_n)$ is Cauchy, there is a $N_1 > 0$, such that for all $n,m > N_1$, $|a_n - a_m| < \epsilon_1$. Similarly, using $(b_n)$ is Cauchy,  we get $N_2$. Let $N = \max(N_1, N_2)$. Then for any $n,m> N$, we have
+$$ |a_n b_n - a_m b_m| =  |a_n b_n - a_n b_m + a_n b_m - a_m b_m| \leq |a_n| |b_n - b_m| + |a_n - a_m| |b_m| \leq M |b_n - b_m| + M |a_n - a_m| \leq M \epsilon_1 + M \epsilon_2 = \epsilon. $$
+=== 5. ===
+We need to prove that, for any $\epsilon>0$, exists $N>0$, such that for all $n>N$, we have
+$$ | a_n b_n - c_n b_n | \leq \epsilon. $$
+Since Cauchy sequences are bounded, there exists $M>0$, such that $|b_n| < M$ for all $n$. Let $\epsilon' = \epsilon / M$, then since $(a_n) \sim (c_n)$, there exists $N>0$, such that for all $n>N$, we have $ |a_n - c_n| < \epsilon'$, hence
+$$ |a_n b_n - c_n b_n | \leq |a_n - c_n| |b_n| \leq |a_n - c_n| M \leq \epsilon' M = \epsilon. $$

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