Lesson 0010: Infinite Limits
Let $(a_k)_{k=m}$ be a sequence of $\Bbb R$. If for every $\alpha\in\Bbb R$, there is a tail of the sequence which is bounded below by $\alpha$, then we write
$$ \lim_{k\to\infty} a_k = \infty$$
If for every $\alpha\in\Bbb R$ there is a tail of the sequence which is bounded above by $\alpha$, then we write
$$ \lim_{k\to\infty} a_k = -\infty $$
In either case, we say that the sequence has an infinite limit.
We define $\Bbb R^* = \Bbb R\cup \{-\infty,\infty\}$ which we call the extended real numbers. We also extend the relation $\le$ by the following rules, for every $x\in \Bbb R^*$.
$$\begin{aligned}
-\infty\le x\le \infty
\end{aligned}$$
We extend the operations of $+$ and $\times$ in the following limited ways. For all $x, y\in\Bbb R$, where $y > 0$,
$$\begin{aligned}
x\pm\infty = y\cdot\pm\infty = \pm\infty, -y\cdot\pm\infty=\mp\infty \\
\infty+\infty = \infty\cdot\infty = -\infty\cdot-\infty = \infty \\
-\infty-\infty = \infty\cdot-\infty = -\infty\cdot \infty = -\infty
\end{aligned}$$
With these defintions, the definition of $\lim_{k\to\infty}a_k$ is now extended to being a value in $\Bbb R^*$. If $\lim_{k\to\infty}a_k = \pm\infty$ then we say either that the limit does not exist as a real number, or that it exists as an extended real number.
To each tail $(a_k)_{k=p}$ we associate with it the infimum of the tail, $\alpha_p = \inf(a_k)_{k=p}$ and the supremum $\beta_p=\sup(a_k)_{k=p}$. We then define the sequence $(\alpha_k)_{k=m}$ which we call the infimum sequence of $(a_k)_{k=p}$. We define $(\beta_k)_{k=p}$ to be the supremum sequence of $(a_k)_{k=m}$.
Then we define the limit infimum of $(a_k)_{k=m}$ to be $\liminf_{k\to\infty} a_k = \lim_{k\to\infty} \alpha_k$, and the limit supremum of $(a_k)_{k=m}$ is $\limsup_{k\to\infty} a_k = \lim_{k\to\infty}\beta_k$.
Throughout this section, assume that $(a_k)_{k=m}$ is a sequence of $\Bbb R$.
Theorem: Infin. lim. no lim.
If the limit is infinite, then the sequence has no real limit.
Statement and proof.
Statement: If $\lim_{k\to\infty}a_k=\pm\infty$ then $(a_k)_{k=m}$ has no limit.
Proof: Suppose $\lim_{k\to\infty} a_k = \infty$ and let $L\in\Bbb R$.
Then let the $p$th tail be bounded below by $L+1$.
We must then have that no tail can be within 1 of $L$, and therefore $\lim_{k\to\infty}a_k\ne L$.
Since $L$ was arbitrary, then no real number is the limit of $(a_k)_{k=m}$.
The proof is mutatis mutandis the same if $\lim_{k\to\infty}a_k=-\infty$.
$\Box$
Theorem: Infin. lim. liminf. incr.
The infimum sequence is monotonically increasing. The supremum sequence is monotonically decreasing.
Statement and proof.
Statement: Let $(\alpha_k)_{k=m}$ be the infimum sequence of $(a_k)_{k=m}$. Then $(\alpha_k)_{k=m}$ is monotonically increasing. The supremum sequence $(\beta_k)_{k=m}$ is monotonically decreasing.
Proof: Let $m\le p < q$. Since $\text{Im}(a_k)_{k=q}\subseteq \text{Im}(a_k)_{k=p}$ then
$$\begin{aligned}
\inf(\text{Im}(a_k)_{k=p}) &\le \inf(\text{Im}(a_k)_{k=q})\\
\sup(\text{Im}(a_k)_{k=p})&\le \sup(\text{Im}_{k=q})
\end{aligned}$$
Theorem: Infin. lim. liminf. infin.
If the liminf is $\infty$ then the limit is $\infty$.
Statement and proof.
Statement: If $\liminf_{k\to\infty} a_k = \infty$ then $\lim_{k\to\infty} a_k = \infty$.
Proof: Let $(\alpha_k)_{k=m}$ be the infimum sequence of $(a_k)_{k=m}$, so $\lim_{k\to\infty}\alpha_k=\infty$.
Let $\alpha\in\Bbb R$ and set $p$ such that the $p$th tail $(\alpha_k)_{k=p}$ is bounded below by $\alpha$.
Then $\alpha\le\alpha_p$ and $\alpha_p$ is a lower bound on $(a_k)_{k=p}$. Therefore $(a_k)_{k=p}$ is bounded below by $\alpha$.
So $\lim_{k\to\infty}a_k = \infty$.
$\Box$
Theorem: Infin. lim. alg.
Products for infinite and null limits.
Statement and proof.
Statement:
- Suppose $(a_k)_{k=m}$ is bounded and $b_k\to 0$. Then $a_kb_k\to 0$.
- Suppose $\liminf_{k\to\infty} a_k > 0$, and that $b_k\to\infty$. Then $a_kb_k \to \infty$.
Proof:
$(a_k)$ bounded, $b_k\to 0$.
Suppose $\alpha$ bounds $(a_k)_{k=m}$ and $b_k\to 0$. Let $\varepsilon\in\Bbb R^+$ and let the $p$th tail of $(b_k)_{k=p}$ be within $\varepsilon/\alpha$ of $0$.
Then for all $k\ge p$,
$$ \begin{aligned}
|a_kb_k| &= |a_k||b_k| \\
& < \alpha(\varepsilon/\alpha) \\
&= \varepsilon
\end{aligned} $$
$\liminf_{k\to\infty}a_k > 0$ and $b_k\to\infty$.
Suppose $\liminf_{k\to\infty} a_k > 0$ and $b_k\to\infty$. Let $(a_k)_{k=p}$ be greater than some $\varepsilon\in\Bbb R^+$.
Let $\alpha\in\Bbb R^+$ and let $(b_k)_{k=q}$ be bounded below by $\alpha/\varepsilon$. Then for every $k\ge \max\{p,q\}$,
$$\begin{aligned}
|a_kb_k| &= |a_k||b_k| \\
&> \varepsilon(\alpha/\varepsilon) \\
&= \alpha
\end{aligned} $$
Therefore $a_kb_k\to\infty$.
Theorem: Infin. lim. recip.
Reciprocals of zero and infinity.
Statement and proof.
Statement: Suppose $0 < a_k$. Then $\lim_{k\to\infty} a_k = 0$ if and only if $\lim_{k\to\infty} \frac{1}{a_k}=\infty$.
Proof:
If $\lim_{k\to\infty} a_k = 0$ then $\lim_{k\to\infty} \frac{1}{a_k}=\infty$.
Suppose $\lim_{k\to\infty}a_k = 0$ and let $\alpha\in\Bbb R^+$. Then there is some tail $(a_k)_{k=p}$ which is less than $\frac{1}{\alpha}$.
That is to say, for every $k\ge p$, we have $a_k < \frac 1 \alpha$. Since $0 < a_k$, we have $\alpha < \frac 1 {a_k}$.
But this shows that $\lim_{k\to\infty} \frac{1}{a_k} = \infty$.
If $\lim_{k\to\infty}\frac{1}{a_k} = \infty$ then $\lim_{k\to\infty} a_k = 0$.
Suppose $\lim_{k\to\infty}\frac{1}{a_k} = \infty$ and let $\varepsilon\in\Bbb R^+$. Then there is some tail $\left(\frac{1}{a_k}\right)_{k=p}$ which is greater than $\frac 1 \varepsilon$.
This means, for $k\ge p$, that $\frac{1}{a_k} > \frac 1 \varepsilon$ and therefore $a_k < \varepsilon$.
Since $0 < a_k$ then $|a_k|<\varepsilon$. Therefore $\lim_{k\to\infty} a_k = 0$.
$\Box$