Tutorial note 4 (limit inferior and superior)

Remark. Whether we use $\limsn,\limin$ or $\dis\limsup_{n\to\infty},\liminf_{n\to\infty}$ is just a matter of taste. In the tutorial notes I will use $\lims,\limi$ (partly because this is the first notation I see when I learn this concept).

Remark. Note that Theorem 6 is a direct consequence of the definition of $\lims$ and $\limi$, we don't need to seek help from $M_k$/$m_k$ Theorem. To see this, suppose that given $r\in \R$ and $\{a_n\}$ a sequence such that \[
\lims a_n<r,
\] then can there be infinitely $a_n$'s such that $a_n\in [r,\infty)$? NO! Indeed, if this happens, then we can extract a subsequence which diverges to $+\infty$ or converges to $\ell\ge r$, either case contradicts to $\lims a_n<r$.

Remark. Theorem 6 actually enables us to characterizes $\lims a_n$ for any bounded sequence $\{a_n\}$ in the following way: \[
\lims a_n=\min \underbrace{\{r\in \R:\forall \epsilon>0, a_n<r+\epsilon\quad\text{for all $n$ big enough}\}}_{:=\dis \mathcal S}.
\tag*{(*)}
\] Indeed, $r=\lims a_n$ is really such a number by theorem 6, i.e., $\lims a_n\in \mathcal S$. To see this, for any $\epsilon>0$, we have \[
\lims a_n<\lims a_n+\epsilon,
\] therefore by Theorem 6 there is an $N$ such that $n>N$ $\implies$ $a_n<\lims a_n+\epsilon$, thus $\lims a_n\in \mathcal S$. On the other hand, suppose $r\in \mathcal S$, then for any $\varepsilon>0$, there is an $N$ such that \[n>N\implies a_n<r+\varepsilon,\] we take $\lims $ on both sides to obtain $\lims a_n\leq r+\varepsilon$. As this is true for each $\varepsilon>0$, we take $\varepsilon\to 0^+$ to conclude $\lims a_n\leq r$, so $\lims a_n = \min \mathcal S$. Therefore we have another definition of $\lims$ as in $(*)$.

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Terminology. Given a sequence $\{a_n\}$ we define the $\mathcal L$ set of this sequence by  \[
\mathcal L = \{a\in [-\infty,\infty]:a_{n_k}\to a \text{ for some subseq $\{a_{n_k}\}$}\}
\]
Example 1. We observe that \[
\{x_n\}=\left\{\frac{\sqrt{3}}{2},\frac{\sqrt{3}}{2},0,-\frac{\sqrt{3}}{2},-\frac{\sqrt{3}}{2},0,\frac{\sqrt{3}}{2},\frac{\sqrt{3}}{2},\dots \right\}.
\] Therefore it is easy to see that all subsequential limits are \[
\mathcal L = \left\{-\frac{\sqrt{3}}{2},0,\frac{\sqrt{3}}{2}\right\},
\] and therefore $\lims x_n=\frac{\sqrt{3}}{2}$ and $\limi x_n=-\frac{\sqrt{3}}{2}$.

Remark. In general it is not easy to compute the $\mathcal L$ set of a given sequence. For example, it is known from Math2031 that:

The $\mathcal L$ set of $\{\sin n:n=1,2,\dots\}$ is precisely $[-1,1]$. 

Although this is not a hard result (indeed quite straightforward if one has the concept of discrete group in $(\R,+)$), it is of course nontrivial, this prompts us to work with the ``biggest'' and ``smallest'' element in $\mathcal L$ in general.

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Example 2.  We first study how $a_n$'s oscillate: \[
\cos \frac{2n\pi}{3}=\left\{ -\frac{1}{2},-\frac{1}{2},1, -\frac{1}{2},-\frac{1}{2},1,\dots \right\}.
\] Therefore it is natural to divide the sequences $\{a_n\}$ into groups of $3$. Indeed, we have \[
\{a_n\}=\{-,-,+,-,-,+,-,-,+,\dots \}
\] Thus it is easy to see that by cancelling countably many negative terms,\[
M_{3n}=\sup\{a_{3n},a_{3n+3},\dots \}=\sup\left\{\frac{(3n)^2}{1+(3n)^2}\right\}=\limn \frac{(3n)^2}{1+(3n)^2} = 1 .
\] Since $\{M_{n}\}$ is decreasing, we have \[
\lims a_n=\limn M_n=\limn M_{3n} = \limn 1 = 1.
\] Next we focus on negative terms to compute $\limi$. Now we have by cancelling countably many positive terms: \[
m_{3n+1} = \inf\{a_{3n+1},a_{3n+2}, \quad a_{3n+4},a_{3n+5},\quad \dots \}.
\] Compare \[
a_{3n+1} =\frac{(3n+1)^2}{1+(3n+1)^2}\brac{-\frac{1}{2}}\quad\text{and}\quad a_{3n+2} =\frac{(3n+2)^2}{1+(3n+2)^2}\brac{-\frac{1}{2}},
\] as the larger the magnitude, the smaller the value, thus we discard $a_{3n+1}$ in each group to conclude \[
m_{3n+1}=\inf\{a_{3n+2},a_{3n+5},a_{3n+8},\dots\} = \limn \frac{(3n+2)^2}{1+(3n+2)^2}\brac{-\frac{1}{2}}=-\frac{1}{2}.
\] Since $\{m_n\}$ is increasing, \[\limi a_n=\limn m_n=\limn m_{3n+1} = -\frac{1}{2}.\]

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Example 3. We first prove the inequality $\lims x_n\leq \lims y_n$. If $\lims y_n=\infty$, then we are done.  Suppose now $\lims y_n<\infty$. If we fix an $\epsilon>0$, then \[\lims y_n<\boxed{\lims y_n+\epsilon,}\] hence by Theorem 6 there is an $N\in \N$ such that \[
n>N \implies y_n<\lims y_n+\epsilon\implies x_n<\lims y_n+\epsilon.\] Recall that $\lims x_{n}$ is a subsequential limit, there is a subsequence $\{x_{n_k}\}$ of $\{x_n\}$ such that $\lim_{k\to\infty} x_{n_k}=\lims x_n$. Now \[\text{$k$ big}\implies x_{n_k}<\lims y_n+\epsilon.\] By taking $k\to\infty$, \[\lims x_n\leq \lims y_n+\epsilon.\] Since this is true for each $\epsilon>0$, by letting $\epsilon\to 0^+$, we have  $\lims x_n\leq \lims y_n$.

Of course the second inequality can be proved directly using the technique in the first paragraph, what we will do is: deduce the second inequality involving $\limi$ from what we have proved. Since $-x_n\ge -y_n$, by taking $\lims$ on both sides, we have \[
\lims(-x_n)\ge \lims(-y_n).
\] By (iii) of Theorem 4 we have \[
-\limi x_n\ge -\limi y_n,
\] therefore we have $\limi x_n\leq \limi y_n$.

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Example 4. There are two ways to show $\limi ka_k=0$.

Method 1. We prove by contradiction. Suppose  $\limi ka_k>0$,  then there is $\alpha>0$ such that $\limi ka_k>\alpha>0$. By Theorem 6 there is a $K\in \N$ such that \[k>K \implies ka_k>\alpha,\] and this implies $a_k>\frac{\alpha}{k}$, so if we sum both sides from $K+1$ to $\infty$, \[\sum_{k=K+1}^\infty a_k\ge \alpha \sum_{k=K+1}^\infty \frac{1}{k},\] this implies $\sum_{k=1}^\infty a_k$ diverges, a contradiction to the hypothesis.


Method 2. Let's fix a $k$ and let 
\[m_k=\inf\{ka_k,(k+1)a_{k+1},\dots\},\] then for each $n\ge k$,  $m_k\leq na_n\implies \frac{m_k}{n}\leq a_n$, by summing from $n=k$ to $n=\infty$,  \[m_k\sum_{n=k}^\infty \frac{1}{n}\leq \sum_{n=k}^\infty a_n, \] since $\sum_{n=1}^\infty a_n$ is finite and $\sum_{n=1}^\infty \frac{1}{n}$ diverges, necessarily $m_k=0$. As this holds for each $k$, \[\limi ka_k=\lim m_k= 0.\]
Finally, we need to raise an example that:

1. $\limi ka_k=0$: For this, we introduce infinitely many $0$ to $a_k$'s since $a_k\ge 0$.
2. $\lims ka_k=1$: For this, we introduce infinitely many $1$ to $ka_k$'s  and keep $ka_k\leq 1$. For instance, we let infinitely many $k$ be s.t. $a_k=1/k$, and set $a_k=0$ otherwise, then the $\mathcal L$ set can only be $\{0,1\}$.
3. $\sum a_k<\infty$: For this, our $k$'s s.t. $a_k=1/k$ should be far enough from each other. 

One possible choice to fulfil 2. and 3. is to let \[
a_k = \begin{cases}
1/k,&k=n^2\text{ for some $n\ge 1$,}\\
0,&\text{otherwise.}
\end{cases}
\] Of course 1. is also satisfied.