1. If $x \in \R$ and there is a Cauchy sequence $(a_n)$ in $\Q$ such that $x = LIM a_n$, then show that $x = \lim a_n$.

2. For any $a \in \R$, prove that $\lim a^n / n! = 0$.

3. Let $A = \{ q \in \Q: q^2 < 3 \}$, let $x = \sup(A)$, prove that $x^2 = 3$.

4. Let $s_1 = 1$, and $s_{n+1} = \sqrt{1 + s_n}$. Assume that $s_n$ converges to $c$, show that $c=(\sqrt{5}+1)/2$.

5. 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.

1. 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 $N = N_1$.

2. 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.

3. 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 $x^2 = 3$.

4. 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 $c = ( \sqrt{5}+1)/2$

5. 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.