More about series following Rudin

• Thm 3.39: radius of convergence for power series
• Summation by part.Thm 3.41, Thm 3.42, Thm 3.44
• Rearrangement theorem of Riemann

Discussion:

• Rudin's exercises in Ch 3, 6,7,9,11

• more about liminf and limsup. Ross 12
• Series. basic definition and examples. partial sum. Cauchy conditions. absolute convergence and convergence. root test, ratio test, alternating series test.

Discussion:

• Ross Ex 12.10, 12.12, 12.13
• Ross Ex 14.1, 14.2, 14.10

Last time, we ended at discussion of two equivalent definitions of subsequential limit. I hope the Cantor's diagonalization trick was fun. Today, we prove the following results

• If $s_n$ converge to $s$, then every subsequence of $s_n$ converges to $s$.
• Every sequence has a monotone subsequence.
  • If there are infinitely many $s_n$, that is 'larger than its tails' ($s_n \geq s_k$ for all $k \geq n$), then we can take the subsequences of such $s_n$, it is monotone decreasing.
  • Otherwise, you throw away an initial chunk that contains such $s_n$, then you can always build an increasing sequence (nothing can stop you)
• Every bounded sequence has convergent subsequence. (just take a monotone sequence, then it will be convergent)
• $\limsup$ and $\liminf$ can be realized as subseq limits.
• Let $(s_n)$ be a seq, and $S$ denote the set of all subseq limits. Then, $S$ is non-empty. $\sup S = \limsup s_n$ and $\inf S = \liminf s_n$. $lim s_n$ exists if and only if $S$ contains only one element. (All these are just summary of previously proven results)
• $S$ is closed under taking limits. (i.e. $S$ is a closed set)
  • Proof of this is fun. Suppose one has a sequence of $t_n$ in $S$, and $t_n \to t$. Can we show that $t$ is in $S$ as well? Well, we need to show that for any $\epsilon>0$, there are infinitely many $s_n$ in $(t-\epsilon, t+\epsilon$. First, we find a $t_n$, such that $|t_n - t| < \epsilon / 2$, then we know there are infinitely many $s_m$ with $|t_n - s_m| < \epsilon/2$, thus these same set of $s_m$ will be $\epsilon$-close to $t$.

Discussion time: Ross 11.2, 11.3, 11.5.

• Cauchy sequence: “we don't know where we are heading, but we agree with each other more and more”.
• Cauchy sequence is equivalent to convergent sequences…. in $\R$.
• Limit sets and subsequences.

First, the definition. Let $(a_n)$ be a sequence in $\R$, we say $(a_n)$ is Cauchy, if for any $\epsilon> 0 $, we have $N>0$, such that for all $n,m>N$, we have $|a_n - a_m | < \epsilon$.

Lemma : Convergent sequence is Cauchy.
Pf: suppose $a_n \to a$. We need to show that for any $\epsilon> 0 $, we have $N>0$, such that for all $n,m>N$, we have $|a_n - a_m | < \epsilon$. Let $\epsilon_1 = \epsilon /2 $, then by convergence of $a_n$, we have $N_1>0$, such that $|a_n - a| < \epsilon_1$ for all $n > N_1$. Thus for any $n,m > N_1$, we have $$ |a_n - a_m| \leq |a_n -a | + |a_m - a| \leq \epsilon_1 + \epsilon_1 = \epsilon $$ Thus, let $N=N_1$ would work.

Lemma : If $A_n \geq a_n \geq B_n$, and $\lim A_n = \lim B_n = a$, then $\lim a_n = a$.
Pf: for any $\epsilon > 0$, we have $N>0$ such that, for all $n > N$, $|A_n - a| < \epsilon, |B_n - a| < \epsilon$, then $$ a_n \leq A_n \leq a+\epsilon, \quad a_n \geq B_n \geq a-\epsilon $$ Thus ,$|a_n - a| \leq \epsilon$ for all $ n > N$, hence $a_n \to a$.

Lemma : Let $(a_n)$ be a bounded sequence. $(a_n)$ is convergent if and only if $\limsup a_n = \liminf a_n$.
Pf: Assume $\limsup a_n = \liminf a_n = a$. Let $A_n = \sup_{m \geq n} a_m, B_n = \inf_{m \geq n} a_m$, then $A_n \geq a_n \geq B_n$. By preview lemma, we know $a_n \to a$.

Assume $(a_n)$ is convergent to $a$. Then, for any $\epsilon > 0$, there is an $N>0$, that for all $n > N$, $|a_n - a|<\epsilon$. In particular, we know $\limsup a_n \leq a+\epsilon$. Since this is true for all $\epsilon>0$, we have $\limsup a_n \leq a$. Similarly, $\liminf a_n \geq a$. On the other hand, since $A_n \geq B_n$, we have $\limsup a_n = \lim A_n \geq \lim B_n = \liminf a_n$. Hence, $\limsup a_n = \liminf a_n = a$.

Lemma: Cauchy sequence is bounded. (Exercise)

Lemma: If $\limsup a_n = A$, then for any $\epsilon > 0$, and any $N > 0$, there exists $n>N$ with $|a_n - A| < \epsilon$. Similarly, if $\limsup a_n = B$, then for any $\epsilon > 0$, and any $N > 0$, there exists $n>N$ with $|a_n - B| < \epsilon$.
Pf: Since $A_n = \sup_{m>n} a_m$ is a monotone decreasing sequence with limit $A$, then for any $\epsilon>0$ and $N>0$, we can find $M > N$ such that $A+\epsilon/2 > A_{M} \geq A$. By definition of $A_{M}$, there exists some $a_n$ with $n \geq M$, that $A_{M} \geq a_n > A_{M} - \epsilon/2$. Now, we have $$ |a_n - A| \leq |a_n - A_M| + |A_M - A| \leq \epsilon/2 + \epsilon/2 = \epsilon $$ The result is proven.

Theorem : Cauchy sequence in $\R$ is convergent.
Proof: Let $(a_n)$ be a Cauchy sequence. By previous lemma, it is bounded. Let $A = \limsup a_n$ and $B = \liminf a_n$, we only need to show that $A = B$ to show $\lim a_n$ exists. We know $A \geq B$ for all bounded sequence $a_n$, suppose $A > B$, and let $\epsilon = (A-B)/3$. Then, by Cauchyness of $(a_n)$, we there is an $N>0$, such that for all $n,m> N$, we have $|a_n - a_m| < \epsilon$. By previous lemma, there exists $n>N$, with $|a_n - A|<\epsilon$, and $m>N$ with $|a_m - B| < \epsilon$. Hence $$ |A - B| \leq |A-a_n|+|a_n - a_m| + |a_m - B| < 3 \epsilon = |A-B|, $$ notice the inequality is strict, hence we have a contradiction. Thus $A = B$.

If $n_1 < n_2 < \cdots $ is a strictly increasing sequence in $\N$, and $(a_n)$ is a sequence, then $(a_{n_k})$ is a subsequence of $(a_n)$.

In this section, we can ask, even if $(a_n)$ itself does not converge to some $a \in \R$, can we cherry-picking some nice subsequence in $(a_n)$ that does converge.

Example:

• $a_n = (-1)^n + 1/n$ has a convergent subsequences, that converges to $+1$; and another subseq convergent to $-1$.

Definition: Let $(a_n)$ be a sequence in $\R$, if $a \in \R$ is the limit of a sequence of $(a_n)$, we say $a$ is a subsequential limit.

Lemma : $a$ is a subsequential limit of $(a_n)$ $\Leftrightarrow$ for any $\epsilon>0, N>0$, there exists an $n>N$, with $|a_n - a| < \epsilon$. Equivalently, for any $\epsilon>0$, the set $A_\epsilon = \{n \mid |a_n - a| < \epsilon \}$ is infinite.

Pf: $\Rightarrow$ : let $a_{n_k}$ be a subseq that converges to $n$, then we can find among member within this subsequence.

$ \Lefgarrow$: it's a good opportunity to introduce the Cantor's diagonal trick. For any positive integer $k$, we know $A_{1/k}$ is infinite, we write it in the $k$-th row, as $$ A_{1/k} = n_{k,1} < n_{k,2}< \cdots $$ As $k=1,2,\cdots$, we have an semi-infinite matrix of indices. We may check that $n_{k, i } \leq n_{k+1, i}$. Thus $n_{k,k} \leq n_{k+1, k} < n_{k+1, k+1}$, hence along the diagonal, we have strictly increasing sequence. Consider the subsequence $a_{n_{kk}}$, that will converge to $a$.

This time we consider some abstract results as what sequences converges.

• monotone bounded sequence converges.
• what is $\limsup$ and $\liminf$? The $\epsilon$-of-room trick.
• Cauchy sequence.

Discussion time: Ex 10.1, 10.7, 10.8 in Ross

This is in Definition 9.8. Informally, it says for any $M$ (no matter how large it is), the sequence will eventually stay above $M$.

A sequence that is (weakly) increasing, namely $a_{n+1} \geq a_n$. Examples: (a) $a_n = n$. (b) $a_n = 1 - n$. © $a_n = \log n$.

Theorem: bounded monotone sequence is convergent.

Just consider the increasing case, the decreasing case is similar. How do we construct such a limit? The limit is obviously the 'largest' number, but there is no largest number in the sequence. We should use 'sup' to get the largest number. But sup needs to applied to a set, not a sequence, so we need to first convert a sequence to its 'value set', $S = \{a_n \mid n \in \N\}$, and define $a = \sup S$. How do we show that $a_n \to a$? We need to show that for any $\epsilon > 0$, we have $|a_n - a| < \epsilon$ for all large enough $n$. Let's first show that, there exists one $n_0$, such that $|a_{n_0} - a| < \epsilon$, this holds by definition of the $\sup S$, namely, there is one element in $S$ that satisfies this condition, it must be $a_{n_0}$ for some $n_0$. Then, we can use monotone condition to say, for all $n > n_0$, we have $a_n \geq a_{n_0}$. Furthermore, since $a_n \leq a$, we know $a_n \in [a_{n_0} , a) \In (a-\epsilon, a)$ for all $n > n_0$, then we are done.

Let's look at the example 2 in Ross. Which is about recursive sequence $$ s_n = \frac{s_{n-1}^2 + 5}{2 s_{n-1}} $$ (such is an example of deterministic 'evolution', how about a probabilist one? That would be a Markov Chain) Let's do the graphical trick. One draw the graph of $y = \frac{x^2+5}{2x}$ on the plane, and the diagonal $y = x$. Then, we can start from point $(x,y) = (s_1, s_1)$. Then, we

• go vertically to the graph, and we land on $(s_1, s_2)$ (since we go vertically, $x$ coordinate is fixed, and we move $y$ until we have $y = f(x)$)
• go horizontally to the diagonal, and we land on $(s_2, s_2)$
• go vertically to the graph, and we land on $(s_2, s_3)$
• ….

This would show a zig-zag curve that goes to the intersection point of $y=x$ and $y=f(x)$, which is point $(s,s)$, where $s$ is the solution to equation $x=f(x)$.

Now, to show that this converges, we apply the above theorem. Note, depending on initial condition, this will either be monotone increasing or decreasing.

Unbounded monotone increasing (respectively, decreasing) sequence converges to $+\infty$ (resp. $-\infty$)

Recall the definition of $\sup$. For $S$ a subset of $\R$, bounded above, we define $\sup(S)$ to be the real number $a$, such that $a$ is $\geq$ than any element in $S$, and for any $\epsilon>0$, there is some $s \in S$ such that $s > a-\epsilon$.

Also, for a sequence $(a_n)_{n=m}^\infty$, we can define the 'value set' $\{a_n\}_{n=m}^\infty$, which is the 'foot print' of the 'journey'.

Also, for a sequence $(a_n)_{n=1}^\infty$, we can define the tail $(a_n)_{n=N}^\infty$, and we only care about the tail of a sequence.

We want to define a gadget, that captures the 'upper envelope' of a sequence, what does that mean? Let $(a_n)$ be a seq, we want define first an auxillary sequence $$ A_m = \sup_{n \geq m} a_n $$ then we define $$ \limsup a_n = \lim A_m (= \inf A_m) $$

Time for some examples, $a_n = (-1)^n (1/n)$.