Math for Non-Geeks/Divergence to infinity/Rules

{{#invoke:Math for Non-Geeks/Seite|oben}} In the last article, we mentioned that some rules for limit calculation carry through to improper convergent sequences, and some don't. For instance, if a sequence $(a_n)$ (improperly) converges to $\mathrm{\infty }$ and a second sequence $(b_n)$ (properly) converges to $0$, one cannot make any statement about the convergence of their product! Math for Non-Geeks: Template:Aufgabe

Rules for computing limits of improperly converging sequences

Which calculation rules for limits of convergent sequences can be carried over to improper convergence? The answer is: almost all of them, but only if certain conditions hold!

Product rule

Suppose that $(a_n)$ is a sequence with $\underset{n\to \mathrm{\infty }}{\lim }{a}_n=\mathrm{\infty }$ . What will happen to the product $(a_n{b}_n)$ ? The case $\underset{n\to \mathrm{\infty }}{\lim }{b}_n=0$ definitely causes trouble, meaning that we cannot make any statement about convergence or divergence of the product.

Case 1: $\underset{n\to \mathrm{\infty }}{\lim }{b}_n=\mathrm{\infty }$. Intuitively, $\mathrm{\infty }\cdot \mathrm{\infty }=\mathrm{\infty }$ so we expect $\underset{n\to \mathrm{\infty }}{\lim }{a}_n{b}_n=\mathrm{\infty }$ . This assertion only needs to be mathematically proven: Template:Math Let $S\in ℝ$ be given. Since $\underset{n\to \mathrm{\infty }}{\lim }{a}_n=\mathrm{\infty }$ we can find an $N_1\in \mathbb{N}$ with $a_n\ge \sqrt{|S|}$ for all $n\ge N_1$. Analogously, since $\underset{n\to \mathrm{\infty }}{\lim }{b}_n=\mathrm{\infty }$ there is an $N_2\in \mathbb{N}$ with $b_n\ge \sqrt{|S|}$ for all $n\ge N_2$. Whenever $n\ge N=\max \{N_1,N_2\}$ we therefore have Template:Math So, indeed $\underset{n\to \mathrm{\infty }}{\lim }{a}_n{b}_n=\mathrm{\infty }$.

Case 2: $\underset{n\to \mathrm{\infty }}{\lim }{b}_n=-\mathrm{\infty }$. Intuitively, $\mathrm{\infty }\cdot -\mathrm{\infty }=-\mathrm{\infty }$ so we expect $\underset{n\to \mathrm{\infty }}{\lim }{a}_n{b}_n=-\mathrm{\infty }$ . What we need to show for a mathematical proof is: Template:Math So let again $S\in ℝ$ be given. Since $\underset{n\to \mathrm{\infty }}{\lim }{a}_n=\mathrm{\infty }$ there is an $N_1\in \mathbb{N}$ with $a_n\ge \sqrt{|S|}$ for all $n\ge N_1$. Analogously, since $\underset{n\to \mathrm{\infty }}{\lim }{b}_n=-\mathrm{\infty }$ there is an $N_2\in \mathbb{N}$ with $b_n\le -\sqrt{|S|}$ for all $n\ge N_2$. Now, for all $n\ge N=\max \{N_1,N_2\}$ we have Template:Math And indeed there is $\underset{n\to \mathrm{\infty }}{\lim }{a}_n{b}_n=-\mathrm{\infty }$.

Case 3: $\underset{n\to \mathrm{\infty }}{\lim }{b}_n=b>0$. Intuitively, $-\mathrm{\infty }\cdot -\mathrm{\infty }=\mathrm{\infty }$ so we again make a guess $\underset{n\to \mathrm{\infty }}{\lim }{a}_n{b}_n=\mathrm{\infty }$. The proof could be done as the two examples above. However, this time we will vary it a bit, to make it not too boring: Template:Math Let $S\in ℝ$ be given. Since $\underset{n\to \mathrm{\infty }}{\lim }{b}_n=b$ , for each $ϵ>0$ there is an $N_1\in \mathbb{N}$ with $|b_n-b|<ϵ$ for all $n\ge N_1$. We set $ϵ=\frac{b}{2}$. Then there is $\mathrm{\forall }n\ge N_1$: $|b_n-b|<\frac{b}{2}$, which especially includes $b_n>\frac{b}{2}>0$. Since $\underset{n\to \mathrm{\infty }}{\lim }{a}_n=\mathrm{\infty }$ there is some $N_2\in \mathbb{N}$ with $a_n\ge \frac{|S|}{\frac{b}{2}}=\frac{2|S|}{b}\ge 0$ for all $n\ge N_1$. Now for $n\ge N=\max \{N_1,N_2\}$ we have Template:Math And hence $\underset{n\to \mathrm{\infty }}{\lim }{a}_n{b}_n=\mathrm{\infty }$.

Case 4: $\underset{n\to \mathrm{\infty }}{\lim }{b}_n=b<0$. Here, $\underset{n\to \mathrm{\infty }}{\lim }{a}_n{b}_n=-\mathrm{\infty }$. Math for Non-Geeks: Template:Aufgabe

Those four cases can also be concluded into one statement. We introduce a practical extension of the real numbers: To the set $ℝ$ , we add the elements $\pm \mathrm{\infty }$ which leads to the bigger set $\overline{ℝ}=ℝ\cup \{\pm \mathrm{\infty }\}$. Math for Non-Geeks: Template:Satz

Sum rule

Let again $(a_n)$ be a sequence with $\underset{n\to \mathrm{\infty }}{\lim }{a}_n=\mathrm{\infty }$. What can we say about the limit of a sum $(a_n+b_n)$ ? For finite $b_n$, the limit will stay unchanged, as intuitively $\mathrm{\infty }+c=\mathrm{\infty }$. Similarly $\mathrm{\infty }+\mathrm{\infty }=\mathrm{\infty }$. The critical case is $\underset{n\to \mathrm{\infty }}{\lim }{b}_n=-\mathrm{\infty }$ , as $\mathrm{\infty }-\mathrm{\infty }$ is not well-defined. In fact, this case does not allow for any statement about convergence or divergence of the sum $(a_n+b_n)$. As an example,

• For $a_n=n$ and $b_n=-n$ there is $\underset{n\to \mathrm{\infty }}{\lim }{a}_n+b_n=\underset{n\to \mathrm{\infty }}{\lim }0=0$.
• For $a_n=2n$ and $b_n=-n$ there is $\underset{n\to \mathrm{\infty }}{\lim }{a}_n+b_n=\underset{n\to \mathrm{\infty }}{\lim }n=\mathrm{\infty }$.

We therefore exclude the case $\underset{n\to \mathrm{\infty }}{\lim }{b}_n=-\mathrm{\infty }$ and consider all other cases:

Case 1: $\underset{n\to \mathrm{\infty }}{\lim }{b}_n=\mathrm{\infty }$. We expect $\underset{n\to \mathrm{\infty }}{\lim }{a}_n+b_n=\mathrm{\infty }$ Mathematically, we need to prove: Template:Math Let $S\in ℝ$ be given. Since $\underset{n\to \mathrm{\infty }}{\lim }{a}_n=\mathrm{\infty }$ there is an $N_1\in \mathbb{N}$ with $a_n\ge \frac{S}{2}$ for all $n\ge N_1$. Analogously, since $\underset{n\to \mathrm{\infty }}{\lim }{b}_n=\mathrm{\infty }$ there is an $N_2\in \mathbb{N}$ with $b_n\ge \frac{S}{2}$ for all $n\ge N_2$. Hence, for all $n\ge N=\max \{N_1,N_2\}$ we have

And indeed $\underset{n\to \mathrm{\infty }}{\lim }{a}_n+b_n=\mathrm{\infty }$.

Case 2: $\underset{n\to \mathrm{\infty }}{\lim }{b}_n=b\in ℝ$. We also expect $\underset{n\to \mathrm{\infty }}{\lim }{a}_n+b_n=\mathrm{\infty }$. Mathematically, we need to prove: Template:Math Let $S\in ℝ$ be given. Since $\underset{n\to \mathrm{\infty }}{\lim }{b}_n=b$ for each $ϵ>0$ we can find an $N_2\in \mathbb{N}$ with $|b_n-b|<ϵ$ for all $n\ge N_2$. This includes the case $ϵ=1$. Hence, $b_n>b-1$ for all $n\ge N_2$. Since $\underset{n\to \mathrm{\infty }}{\lim }{a}_n=\mathrm{\infty }$ there is also an $N_1\in \mathbb{N}$ with $a_n\ge S-b+1$ for all $n\ge N_1$. Hence, for any $n\ge N=\max \{N_1,N_2\}$ we have

And we get the desired result $\underset{n\to \mathrm{\infty }}{\lim }{a}_n+b_n=\mathrm{\infty }$.

Both cases can be concluded in a theorem:

Inversion

This rule is also quite intuitive: Let $(a_n)$ be a sequence with $a_n\ne 0$ for all $n\in \mathbb{N}$ and $\underset{n\to \mathrm{\infty }}{\lim }{a}_n=\mathrm{\infty }$ or $\underset{n\to \mathrm{\infty }}{\lim }{a}_n=-\mathrm{\infty }$, then $\left(\frac{1}{a_n}\right)$ formally converges to $\frac{1}{\mathrm{\infty }}=0$ and should hence be a null sequence. Is this really mathematically true?

Case 1: $\underset{n\to \mathrm{\infty }}{\lim }{a}_n=\mathrm{\infty }$. We need to show Template:Math Let $ϵ>0$ be given. Since $\underset{n\to \mathrm{\infty }}{\lim }{a}_n=\mathrm{\infty }$ for any $S=\frac{1}{ϵ}+1$ there is an $N\in \mathbb{N}$, such that $\mathrm{\forall }n\ge N$ there is $a_n\ge \frac{1}{ϵ}+1$. Hence, $\mathrm{\forall }n\ge N$ there is Template:Math So$\underset{n\to \mathrm{\infty }}{\lim }{a}_n=0$.

Case 2: $\underset{n\to \mathrm{\infty }}{\lim }{a}_n=-\mathrm{\infty }$. Math for Non-Geeks: Template:Aufgabe

We conclude these findings in a theorem: Math for Non-Geeks: Template:Satz

The question is now: can we define a "converse of the inversion rule" which holds under more special assumptions? The sequence $((-1{)}^{n}n)$ is not diverging to $\mathrm{\infty }$ or $-\mathrm{\infty }$ because it keeps changing presign, so there is a subsequence of it converging to $\mathrm{\infty }$ and a subsequence converging to $-\mathrm{\infty }$. We can avoid this by forbidding a change of presign in $(a_n)$. It should also not be too bad if the change of presign is allowed again on finitely many elements, since a manipulation on finitely many elements never changes convergence properties.

Case 1: Let $(a_n)$ be a sequence with $\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{a_n}=0$, all sequence elements being $\ne 0$ and all but finitely many sequence elements being positive. Then, intuitively $\underset{n\to \mathrm{\infty }}{\lim }{a}_n=\frac{1}{+0}=\mathrm{\infty }$. For a mathematical proof, we need to show that Template:Math Let $S\in ℝ$ be given. Since $\left(\frac{1}{a_n}\right)$ is a null sequence, for $ϵ=\frac{1}{|S|}>0$ we can find an $\tilde{N}\in \mathbb{N}$ with $\left|\frac{1}{a_n}\right|<\frac{1}{|S|}$ for all $n\ge \tilde{N}$. Since almost all elements of $(a_n)$ are positive, there is an $N\ge \tilde{N}$ with $\left|\frac{1}{a_n}\right|=\frac{1}{a_n}<\frac{1}{|S|}$ for all $n\ge N$. Therefore $a_n\ge |S|\ge S$ for all $n\ge N$. So we get $\underset{n\to \mathrm{\infty }}{\lim }{a}_n=\mathrm{\infty }$.

Case 2: Let now $(a_n)$ v $\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{a_n}=0$, all sequence elements being $\ne 0$ but this time, almost all of them are negative. Math for Non-Geeks: Template:Aufgabe

The converse of the inversion rule is also concluded into a theorem. Math for Non-Geeks: Template:Satz

Quotient rule

The inversion rule is an example of a quotient $\frac{a_n}{b_n}=\frac{1}{b_n}$ of sequences $(a_n)$ and $(b_n)$ wit constant $a_n=1$. Now, we generalize to quotients $\left(\frac{a_n}{b_n}\right)$ of any sequences $(a_n)$ and $(b_n)$ with $b_n\ne 0$ for all $n\in \mathbb{N}$ .

First, we consider $\underset{n\to \mathrm{\infty }}{\lim }{b}_n=\mathrm{\infty }$ . At this point, we exclude the cases $\underset{n\to \mathrm{\infty }}{\lim }{a}_n=\pm \mathrm{\infty }$ , since $\frac{\pm \mathrm{\infty }}{\mathrm{\infty }}$ is ill-defined. Let $\underset{n\to \mathrm{\infty }}{\lim }{a}_n=a\in ℝ$. Then, formally $\underset{n\to \mathrm{\infty }}{\lim }\frac{a_n}{b_n}=\frac{a}{\mathrm{\infty }}=0$. To verify this mathematically, we need to show Template:Math Let $ϵ>0$ be given. Since there is convergence $\underset{n\to \mathrm{\infty }}{\lim }{a}_n=a$ , the sequence $(a_n)$ must be bounded, i.e. there is a $K>0$ with $|a_n|\le K$ for all $n\in \mathbb{N}$. Now since $\underset{n\to \mathrm{\infty }}{\lim }{b}_n=\mathrm{\infty }$, the inversion rule implies $\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{b_n}=0$. So there is an $N\in \mathbb{N}$ with $\left|\frac{1}{b_n}\right|<\frac{ϵ}{K}$ for all $n\ge N$. Hence, $\mathrm{\forall }n\ge N$ there is: Template:Math And we have convergence $\underset{n\to \mathrm{\infty }}{\lim }\frac{a_n}{b_n}=0$.

The case $\underset{n\to \mathrm{\infty }}{\lim }{b}_n=-\mathrm{\infty }$ and $\underset{n\to \mathrm{\infty }}{\lim }{a}_n=a\in ℝ$ also leads to $\underset{n\to \mathrm{\infty }}{\lim }\frac{a_n}{b_n}=0$ by the same argument.

We conclude Math for Non-Geeks: Template:Satz

Next, we let the enumerator diverge as $\underset{n\to \mathrm{\infty }}{\lim }{a}_n=\mathrm{\infty }$. The case $\underset{n\to \mathrm{\infty }}{\lim }{b}_n=\pm \mathrm{\infty }$ again leads to the ill-defined expression $\frac{\mathrm{\infty }}{\pm \mathrm{\infty }}$ and will not be not considered at this point.

Case 1: $\underset{n\to \mathrm{\infty }}{\lim }{b}_n=b>0$. Here, we assert $\underset{n\to \mathrm{\infty }}{\lim }\frac{a_n}{b_n}=\frac{\mathrm{\infty }}{b}=\mathrm{\infty }$. : Template:Math Let $S\in ℝ$ be given. Since $\left(b_n\right)$ converges to $b>0$, there must be an $N_1\in \mathbb{N}$, such that $b_n\le \frac{3}{2}b$ for all $n\ge N_1$. Since $\underset{n\to \mathrm{\infty }}{\lim }{a}_n=\mathrm{\infty }$ there is also an $N_2\in \mathbb{N}$ with $a_n\ge \frac{3}{2}b|S|$ for all $n\ge N_2$. Hence, for all $n\ge N=\max \{N_1,N_2\}$, there is: Template:Math So we have convergence $\underset{n\to \mathrm{\infty }}{\lim }\frac{a_n}{b_n}=\mathrm{\infty }$.

Case 2: $\underset{n\to \mathrm{\infty }}{\lim }{b}_n=0$ with almost all $b_n$ being positive. Here, we assert $\underset{n\to \mathrm{\infty }}{\lim }\frac{a_n}{b_n}=\frac{\mathrm{\infty }}{+0}=\mathrm{\infty }$. A mathematical proof requires showing Template:Math Let $S\in ℝ$ be given. Since $\left(b_n\right)$ converges to $0$ and almost all elements are positive, there must be an $N_1\in \mathbb{N}$ with $b_n\le \frac{1}{2}$ for all $n\ge N_1$. Since $\underset{n\to \mathrm{\infty }}{\lim }{a}_n=\mathrm{\infty }$ there is also an $N_2\in \mathbb{N}$ with $a_n\ge \frac{1}{2}|S|$ for all $n\ge N_2$. So for all $n\ge N=\max \{N_1,N_2\}$, there is: Template:Math And again, we have convergence $\underset{n\to \mathrm{\infty }}{\lim }\frac{a_n}{b_n}=\mathrm{\infty }$.

All 4 cases are concluded in a theorem Math for Non-Geeks: Template:Satz

Direct comparison

Intuitively, if $(x_n)$ is given and some "smaller" sequence $(y_n)$ diverges to $\mathrm{\infty }$, then also the "bigger" $(x_n)$ must tend to $\mathrm{\infty }$. This should still hold true if "$(x_n)$ is bigger than $(y_n)$" almost everywhere. Mathematically, we need to show Template:Math So let $S\in ℝ$ be given. Since $\underset{n\to \mathrm{\infty }}{\lim }{y}_n=\mathrm{\infty }$ there is an $\tilde{N}\in \mathbb{N}$ with $y_n\ge S$ for all $n\ge \tilde{N}$. Since $x_n\ge y_n$ for all but finitely many $n\in \mathbb{N}$ there is an $N\ge \tilde{N}$ with $x_n\ge y_n\ge S$ for all $n\ge N$. So indeed, $\underset{n\to \mathrm{\infty }}{\lim }{x}_n=\mathrm{\infty }$.

We conclude this in a theorem: Math for Non-Geeks: Template:Satz

Of course, a similar statement holds true for $x_n\le y_n$ and $\underset{n\to \mathrm{\infty }}{\lim }{y}_n=-\mathrm{\infty }$. Then also $\underset{n\to \mathrm{\infty }}{\lim }{x}_n=\mathrm{\infty }$ . This can easily seen by considering the sequences $(|x_n|{)}_{n\in \mathbb{N}}$ and $(|y_n|{)}_{n\in \mathbb{N}}$.

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