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--- math104-f21:midterm1-review [2021/09/17 21:05]
pzhou [Midterm 1: Review]
+++ math104-f21:midterm1-review [2022/01/11 08:36] (current)
pzhou ↷ Page moved from math104:midterm1-review to math104-f21:midterm1-review
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-==== Sample Problems ====
+===== Sample Problems =====
+<del>Midterm will have 3 of such following questions.</del> It turns out 3 problems of the following kinds maybe either too easy or too hard. The number of problems may vary, but the difficulties will include some easy ones and some hard ones.
+=== True or False? If true, prove your result; if false, give a counter example. ===
+  - Let $(a_n)$ be a Cauchy sequence of irrational numbers, then its limit has to be an irrational number.
+  - Let $S$ be an ordered set, then any non-empty finite subset $E \In S$ has a least upper bound.
+  - If $x$ is a limit point of sequence $(a_n)$, then there exists one $n \in \N$, such that $a_n = x$.
+  - If $(a_n)$ is a sequence bounded above, and $L = \limsup(a_n)$, then for any $\epsilon > 0$, there exists an integer $N > 0$, such that $a_n < L + \epsilon$.
+  - Let $(a_n)$ be a bounded sequence in $\R$. Let $A_n = \sup \{a_m : 0 \leq m \leq n \}$, then $\lim A_n = \limsup a_n$
+  - Let $(a_n)$ and $(b_n)$ be convergent sequences with the same limit $x$. And choose any function $f: \N \to \{0,1\}$. We define a new sequence, by mixing $a_n$ and $b_n$ $$c_n = \begin{cases} a_n & \text{if } f(n) = 0 \cr b_n & \text{ if }  f(n) = 1 \end{cases} $$ Then, $c_n$ converges to $x$.
+  - If $(a_n)$ and $(b_n)$ are Cauchy sequence in $\R$, and they satisfy that $\lim(a_n b_n) = 1$, then $\lim a_n \neq 0$.
+  - If $(a_n)$ is a sequence of positive real numbers, for $n \geq 1$,  and $A_n = (a_1 + \cdots + a_n) / n$, show that if $a_n$ is convergent then $A_n$ is convergent. Give an example where $A_n$ is convergent, but $a_n$ is not convergent.
+  - Is there a sequence $(a_n)$, where $|a_n - a_{n-1}|$ is monotone decreasing, but $(a_n)$ is not convergent?
+===== Solution =====
+==== 1 ====
+Let $(a_n)$ be a Cauchy sequence of irrational numbers, then its limit has to be an irrational number.
+False. Say $a_n = \pi / n$.
+==== 2 ====
+Let $S$ be an ordered set, then any non-empty finite subset $E \In S$ has a least upper bound.
+True. One can prove this by induction on the size of $E$.
+==== 3 ====
+If $x$ is a limit point of sequence $(a_n)$, then there exists one $n \in \N$, such that $a_n = x$.
+False. $0 = \lim_n 1/n$.
+==== 4 ====
+If $(a_n)$ is a sequence bounded above, and $L = \limsup(a_n)$, then for any $\epsilon > 0$, there exists an integer $N > 0$, such that $a_n < L + \epsilon$.
+True. explained in class.
+==== 5 ====
+Let $(a_n)$ be a bounded sequence in $\R$. Let $A_n = \sup \{a_m : 0 \leq m \leq n \}$, then $\lim A_n = \limsup a_n$.
+False. That's not the definition of $\limsup$. Say $a_n = 1/(n+1)$, for $n \in \N$, then $\lim A_n = 1$, and $\limsup a_n=0$.
+==== 6 ====
+Let $(a_n)$ and $(b_n)$ be convergent sequences with the same limit $x$. And choose any function $f: \N \to \{0,1\}$. We define a new sequence, by mixing $a_n$ and $b_n$ $$c_n = \begin{cases} a_n & \text{if } f(n) = 0 \cr b_n & \text{ if }  f(n) = 1 \end{cases} $$ Then, $c_n$ converges to $x$.
+True. By convergence of $a_n$ and $b_n$, for any $\epsilon>0$, exists $N_1 > 0$ and $N_2$, such that if $n > N_1$, then $|a_n - x| < \epsilon$, and if $n>N_2$, then $|b_n - x| < \epsilon$. Since $c_n$ is either $a_n$ or $b_n$, hence for $n > N=\max(N_1, N_2)$, we have $|c_n - x| < \epsilon$.
+==== 7 ====
+ If $(a_n)$ and $(b_n)$ are Cauchy sequence in $\R$, and they satisfy that $\lim(a_n b_n) = 1$, then $\lim a_n \neq 0$.
+True. Since if $a_n, b_n$ are convergent, then $1 = \lim(a_n b_n) = (\lim a_n ) (\lim b_n)$, hence $\lim a_n \neq 0$.
+==== 8 ====
+If $(a_n)$ is a sequence of positive real numbers, for $n \geq 1$,  and $A_n = (a_1 + \cdots + a_n) / n$, show that if $a_n$ is convergent then $A_n$ is convergent. Give an example where $A_n$ is convergent, but $a_n$ is not convergent.
+Proof: Let $x$ be the limit of $a_n$. Given any $\epsilon > 0$, we claim there exists an $N>0$, such that $|A_n - x| < \epsilon$ for all $n > N$. Let $\epsilon_1 = \epsilon/3$, and let $N_1>0$ be such that for all $n \geq N_1$, we have $|a_n - x| < \epsilon_1$. Since $a_n$ is convergent, we also have an $M>0$, such that $|a_n| < M$ for all $n$. Then, for any $n > N_1$, we have
+$$ A_n - x = (1/n) \sum_{m=1}^{N_1} a_m + (1/n) \sum_{m=N_1+1}^n (a_m-x) + (N_1/n) x $$
+Thus, we have
+$$ |A_n - x| \leq (1/n) N_1 M + (1 - N_1/n) \epsilon_1 + (N_1/n) x $$
+we may choose $N$ large enough, such that $N_1 / N x < \epsilon/3$, $N_1/N M < \epsilon / 3$, thus, for any $n > N$, we have
+$$ |A_n - x| \leq \epsilon/3 + \epsilon / 3 + \epsilon / 3 = \epsilon. $$
+This finishes the proof of the first claim.
+Here is an example where $A_n$ is convergent, but $a_n$ is not convergent: $a_n = (-1)^n$.
+==== 9 ====
+Is there a sequence $(a_n)$, where $|a_n - a_{n-1}|$ is monotone decreasing, but $(a_n)$ is not convergent?
+Yes, for $n \geq 1$, $a_n = \sum_{m=1}^n 1/n$.

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