Divergence to infinity/Rules – "Math for Non-Geeks"

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 $\infty$ 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 $\lim _{n\to \infty }a_{n}=\infty$ . What will happen to the product $(a_{n}b_{n})$ ? The case $\lim _{n\to \infty }b_{n}=0$ definitely causes trouble, meaning that we cannot make any statement about convergence or divergence of the product.

Case 1: $\lim _{n\to \infty }b_{n}=\infty$. Intuitively, $\infty \cdot \infty =\infty$ so we expect $\lim _{n\to \infty }a_{n}b_{n}=\infty$ . This assertion only needs to be mathematically proven: $\forall S\in \mathbb {R} \ \exists N\in \mathbb {N} \ \forall n\geq N:a_{n}b_{n}\geq S$ Let $S\in \mathbb {R}$ be given. Since $\lim _{n\to \infty }a_{n}=\infty$ we can find an $N_{1}\in \mathbb {N}$ with $a_{n}\geq {\sqrt {|S|}}$ for all $n\geq N_{1}$. Analogously, since $\lim _{n\to \infty }b_{n}=\infty$ there is an $N_{2}\in \mathbb {N}$ with $b_{n}\geq {\sqrt {|S|}}$ for all $n\geq N_{2}$. Whenever $n\geq N=\max\{N_{1},N_{2}\}$ we therefore have $a_{n}b_{n}\geq {\sqrt {|S|}}{\sqrt {|S|}}=|S|\geq S$ So, indeed $\lim _{n\to \infty }a_{n}b_{n}=\infty$.

Case 2: $\lim _{n\to \infty }b_{n}=-\infty$. Intuitively, $\infty \cdot -\infty =-\infty$ so we expect $\lim _{n\to \infty }a_{n}b_{n}=-\infty$ . What we need to show for a mathematical proof is: $\forall S\in \mathbb {R} \ \exists n\in \mathbb {N} \ \forall n\geq N:a_{n}b_{n}\leq S$ So let again $S\in \mathbb {R}$ be given. Since $\lim _{n\to \infty }a_{n}=\infty$ there is an $N_{1}\in \mathbb {N}$ with $a_{n}\geq {\sqrt {|S|}}$ for all $n\geq N_{1}$. Analogously, since $\lim _{n\to \infty }b_{n}=-\infty$ there is an $N_{2}\in \mathbb {N}$ with $b_{n}\leq -{\sqrt {|S|}}$ for all $n\geq N_{2}$. Now, for all $n\geq N=\max\{N_{1},N_{2}\}$ we have $\underbrace {a_{n}} _{\geq {\sqrt {|S|}}\geq 0}\overbrace {b_{n}} ^{\leq -{\sqrt {|S|}}\leq 0}\leq {\sqrt {|S|}}(-{\sqrt {|S|}})=-|S|\leq S$ And indeed there is $\lim _{n\to \infty }a_{n}b_{n}=-\infty$.

Case 3: $\lim _{n\to \infty }b_{n}=b>0$. Intuitively, $-\infty \cdot -\infty =\infty$ so we again make a guess $\lim _{n\to \infty }a_{n}b_{n}=\infty$. 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: $\forall S\in \mathbb {R} \ \exists n\in \mathbb {N} \ \forall n\geq N:a_{n}b_{n}\geq S$ Let $S\in \mathbb {R}$ be given. Since $\lim _{n\to \infty }b_{n}=b$ , for each $\epsilon >0$ there is an $N_{1}\in \mathbb {N}$ with $|b_{n}-b|<\epsilon$ for all $n\geq N_{1}$. We set $\epsilon ={\tfrac {b}{2}}$. Then there is $\forall n\geq N_{1}$: $|b_{n}-b|<{\tfrac {b}{2}}$, which especially includes $b_{n}>{\tfrac {b}{2}}>0$. Since $\lim _{n\to \infty }a_{n}=\infty$ there is some $N_{2}\in \mathbb {N}$ with $a_{n}\geq {\tfrac {|S|}{\tfrac {b}{2}}}={\tfrac {2|S|}{b}}\geq 0$ for all $n\geq N_{1}$. Now for $n\geq N=\max\{N_{1},N_{2}\}$ we have $a_{n}b_{n}\geq {\tfrac {2|S|}{b}}{\tfrac {b}{2}}=|S|\geq S$ And hence $\lim _{n\to \infty }a_{n}b_{n}=\infty$.

Case 4: $\lim _{n\to \infty }b_{n}=b<0$. Here, $\lim _{n\to \infty }a_{n}b_{n}=-\infty$. 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 $\mathbb {R}$ , we add the elements $\pm \infty$ which leads to the bigger set ${\overline {\mathbb {R} }}=\mathbb {R} \cup \{\pm \infty \}$.

Theorem (Product rule for improperly convergent sequences)

Let $(a_{n})_{n\in \mathbb {N} }$ be a real sequence with $\lim _{n\to \infty }a_{n}=\infty$ and $(b_{n})_{n\in \mathbb {N} }$ a real sequence with $\lim _{n\to \infty }b_{n}=b\in {\overline {\mathbb {R} }}$. Then $\lim _{n\to \infty }a_{n}b_{n}={\begin{cases}\infty &{\text{ if }}b>0{\text{ or }}b=\infty ,\\-\infty &{\text{ if }}b<0{\text{ or }}b=-\infty .\end{cases}}$

Sum rule

Let again $(a_{n})$ be a sequence with $\lim _{n\to \infty }a_{n}=\infty$. 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 $\infty +c=\infty$. Similarly $\infty +\infty =\infty$. The critical case is $\lim _{n\to \infty }b_{n}=-\infty$ , as $\infty -\infty$ 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 $\lim _{n\to \infty }a_{n}+b_{n}=\lim _{n\to \infty }0=0$.
• For $a_{n}=2n$ and $b_{n}=-n$ there is $\lim _{n\to \infty }a_{n}+b_{n}=\lim _{n\to \infty }n=\infty$.

We therefore exclude the case $\lim _{n\to \infty }b_{n}=-\infty$ and consider all other cases:

Case 1: $\lim _{n\to \infty }b_{n}=\infty$. We expect $\lim _{n\to \infty }a_{n}+b_{n}=\infty$ Mathematically, we need to prove: $\forall S\in \mathbb {R} \ \exists n\in \mathbb {N} \ \forall n\geq N:a_{n}+b_{n}\geq S$ Let $S\in \mathbb {R}$ be given. Since $\lim _{n\to \infty }a_{n}=\infty$ there is an $N_{1}\in \mathbb {N}$ with $a_{n}\geq {\tfrac {S}{2}}$ for all $n\geq N_{1}$. Analogously, since $\lim _{n\to \infty }b_{n}=\infty$ there is an $N_{2}\in \mathbb {N}$ with $b_{n}\geq {\tfrac {S}{2}}$ for all $n\geq N_{2}$. Hence, for all $n\geq N=\max\{N_{1},N_{2}\}$ we have

$a_{n}+b_{n}\geq {\tfrac {S}{2}}+{\tfrac {S}{2}}=S$

And indeed $\lim _{n\to \infty }a_{n}+b_{n}=\infty$.

Case 2: $\lim _{n\to \infty }b_{n}=b\in \mathbb {R}$. We also expect $\lim _{n\to \infty }a_{n}+b_{n}=\infty$. Mathematically, we need to prove: $\forall S\in \mathbb {R} \ \exists N\in \mathbb {N} \ \forall n\geq N:a_{n}+b_{n}\geq S$ Let $S\in \mathbb {R}$ be given. Since $\lim _{n\to \infty }b_{n}=b$ for each $\epsilon >0$ we can find an $N_{2}\in \mathbb {N}$ with $|b_{n}-b|<\epsilon$ for all $n\geq N_{2}$. This includes the case $\epsilon =1$. Hence, $b_{n}>b-1$ for all $n\geq N_{2}$. Since $\lim _{n\to \infty }a_{n}=\infty$ there is also an $N_{1}\in \mathbb {N}$ with $a_{n}\geq S-b+1$ for all $n\geq N_{1}$. Hence, for any $n\geq N=\max\{N_{1},N_{2}\}$ we have

$a_{n}+b_{n}\geq (S-b+1)+b-1=S$

And we get the desired result $\lim _{n\to \infty }a_{n}+b_{n}=\infty$.

Both cases can be concluded in a theorem:

Theorem (Sum rule for improperly convergent sequences)

Let $(a_{n})_{n\in \mathbb {N} }$ be a real sequence with $\lim _{n\to \infty }a_{n}=\infty$ and $(b_{n})_{n\in \mathbb {N} }$ be a real sequence with $\lim _{n\to \infty }b_{n}=b\in \mathbb {R} \cup \{\infty \}$. Then $\lim _{n\to \infty }a_{n}+b_{n}=\infty$

Inversion

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

Case 1: $\lim _{n\to \infty }a_{n}=\infty$. We need to show $\forall \epsilon >0\ \exists N\in \mathbb {N} \ \forall n\geq N:\left|{\tfrac {1}{a_{n}}}\right|<\epsilon$ Let $\epsilon >0$ be given. Since $\lim _{n\to \infty }a_{n}=\infty$ for any $S={\tfrac {1}{\epsilon }}+1$ there is an $N\in \mathbb {N}$, such that $\forall n\geq N$ there is $a_{n}\geq {\tfrac {1}{\epsilon }}+1$. Hence, $\forall n\geq N$ there is $\left|{\tfrac {1}{a_{n}}}\right|\leq {\tfrac {1}{{\tfrac {1}{\epsilon }}+1}}<{\tfrac {1}{\tfrac {1}{\epsilon }}}=\epsilon$ So$\lim _{n\to \infty }a_{n}=0$.

Case 2: $\lim _{n\to \infty }a_{n}=-\infty$. Math for Non-Geeks: Template:Aufgabe

We conclude these findings in a theorem:

Theorem (Inversion rule 1 for improperly convergent series)

Let $(a_{n})_{n\in \mathbb {N} }$ be a real sequence with $a_{n}\neq 0$ for all $n\in \mathbb {N}$ and $\lim _{n\to \infty }a_{n}=a\in \{-\infty ,\infty \}$. Then $\lim _{n\to \infty }{\tfrac {1}{a_{n}}}=0$

Question: Does the converse of the inversion rule, i.e. if $\lim _{n\to \infty }{\tfrac {1}{a_{n}}}=0$ we may imply $\lim _{n\to \infty }a_{n}=\infty$ or $\lim _{n\to \infty }a_{n}=-\infty$?

Nope! A counterexample is $(a_{n})$ with $a_{n}=(-1)^{n}n$. Then, $\lim _{n\to \infty }{\tfrac {1}{a_{n}}}=\lim _{n\to \infty }{\tfrac {1}{(-1)^{n}n}}={\tfrac {(-1)^{n}}{n}}=0$, but $(a_{n})$ does not go to $\infty$ or $-\infty$. However, it is true that the absolute values $|a_{n}|$ diverge to $\infty$.

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 $\infty$ or $-\infty$ because it keeps changing presign, so there is a subsequence of it converging to $\infty$ and a subsequence converging to $-\infty$. 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 $\lim _{n\to \infty }{\tfrac {1}{a_{n}}}=0$, all sequence elements being $\neq 0$ and all but finitely many sequence elements being positive. Then, intuitively $\lim _{n\to \infty }a_{n}={\frac {1}{+0}}=\infty$. For a mathematical proof, we need to show that $\forall S\in \mathbb {R} \ \exists N\in \mathbb {N} \ \forall n\geq N:a_{n}\geq S$ Let $S\in \mathbb {R}$ be given. Since $\left({\tfrac {1}{a_{n}}}\right)$ is a null sequence, for $\epsilon ={\tfrac {1}{|S|}}>0$ we can find an ${\tilde {N}}\in \mathbb {N}$ with $\left|{\tfrac {1}{a_{n}}}\right|<{\tfrac {1}{|S|}}$ for all $n\geq {\tilde {N}}$. Since almost all elements of $(a_{n})$ are positive, there is an $N\geq {\tilde {N}}$ with $\left|{\tfrac {1}{a_{n}}}\right|={\tfrac {1}{a_{n}}}<{\tfrac {1}{|S|}}$ for all $n\geq N$. Therefore $a_{n}\geq |S|\geq S$ for all $n\geq N$. So we get $\lim _{n\to \infty }a_{n}=\infty$.

Case 2: Let now $(a_{n})$ v $\lim _{n\to \infty }{\tfrac {1}{a_{n}}}=0$, all sequence elements being $\neq 0$ 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.

Theorem (Inversion rule 2 for improperly convergent series)

Let $(a_{n})_{n\in \mathbb {N} }$ be a real sequence with $a_{n}\neq 0$ for all $n\in \mathbb {N}$ and $\lim _{n\to \infty }{\tfrac {1}{a_{n}}}=0$. Then

• $\lim _{n\to \infty }a_{n}=\infty$, if $a_{n}>0$ for almost all $n\in \mathbb {N}$
• $\lim _{n\to \infty }a_{n}=-\infty$, if $a_{n}<0$ for almost all $n\in \mathbb {N}$

Example (Inversion rule)

In the article examples for limits, we have proven that "exponential sequences win against polynomial ones", i.e. $\left({\tfrac {n^{k}}{z^{n}}}\right)_{n\in \mathbb {N} }$ is a null sequence for $|z|>1$ and $k\in \mathbb {N}$. If $z>1$, all sequence elements are non-negative. So the inversion rule implies $\lim _{n\to \infty }{\tfrac {1}{\tfrac {n^{k}}{z^{n}}}}=\lim _{n\to \infty }{\tfrac {z^{n}}{n^{k}}}=\infty$

Analogously, for $z>1$: $\lim _{n\to \infty }{\tfrac {n!}{z^{n}}}=\infty$

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 $\left({\tfrac {a_{n}}{b_{n}}}\right)$ of any sequences $(a_{n})$ and $(b_{n})$ with $b_{n}\neq 0$ for all $n\in \mathbb {N}$ .

First, we consider $\lim _{n\to \infty }b_{n}=\infty$ . At this point, we exclude the cases $\lim _{n\to \infty }a_{n}=\pm \infty$ , since ${\frac {\pm \infty }{\infty }}$ is ill-defined. Let $\lim _{n\to \infty }a_{n}=a\in \mathbb {R}$. Then, formally $\lim _{n\to \infty }{\tfrac {a_{n}}{b_{n}}}={\frac {a}{\infty }}=0$. To verify this mathematically, we need to show $\forall \epsilon >0\ \exists N\in \mathbb {N} \ \forall n\geq N:\left|{\tfrac {a_{n}}{b_{n}}}\right|<\epsilon$ Let $\epsilon >0$ be given. Since there is convergence $\lim _{n\to \infty }a_{n}=a$ , the sequence $(a_{n})$ must be bounded, i.e. there is a $K>0$ with $|a_{n}|\leq K$ for all $n\in \mathbb {N}$. Now since $\lim _{n\to \infty }b_{n}=\infty$, the inversion rule implies $\lim _{n\to \infty }{\tfrac {1}{b_{n}}}=0$. So there is an $N\in \mathbb {N}$ with $\left|{\tfrac {1}{b_{n}}}\right|<{\tfrac {\epsilon }{K}}$ for all $n\geq N$. Hence, $\forall n\geq N$ there is: $\left|{\tfrac {a_{n}}{b_{n}}}\right|=|a_{n}|\cdot \left|{\tfrac {1}{b_{n}}}\right|<K\cdot {\tfrac {\epsilon }{K}}=\epsilon$ And we have convergence $\lim _{n\to \infty }{\tfrac {a_{n}}{b_{n}}}=0$.

The case $\lim _{n\to \infty }b_{n}=-\infty$ and $\lim _{n\to \infty }a_{n}=a\in \mathbb {R}$ also leads to $\lim _{n\to \infty }{\tfrac {a_{n}}{b_{n}}}=0$ by the same argument.

We conclude

Theorem (Quotient rule 1 for improperly convergent sequences)

Let $(a_{n})_{n\in \mathbb {N} }$ and $(b_{n})_{n\in \mathbb {N} }$ be real sequences with $\lim _{n\to \infty }a_{n}=a\in \mathbb {R}$ and $\lim _{n\to \infty }b_{n}=b\in \{-\infty ,\infty \}$. Then $\lim _{n\to \infty }{\tfrac {a_{n}}{b_{n}}}=0$.

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

Case 1: $\lim _{n\to \infty }b_{n}=b>0$. Here, we assert $\lim _{n\to \infty }{\tfrac {a_{n}}{b_{n}}}={\frac {\infty }{b}}=\infty$. : $\forall S\in \mathbb {R} \ \exists N\in \mathbb {N} \ \forall n\geq N:{\tfrac {a_{n}}{b_{n}}}\geq S$ Let $S\in \mathbb {R}$ be given. Since $\left(b_{n}\right)$ converges to $b>0$, there must be an $N_{1}\in \mathbb {N}$, such that $b_{n}\leq {\tfrac {3}{2}}b$ for all $n\geq N_{1}$. Since $\lim _{n\to \infty }a_{n}=\infty$ there is also an $N_{2}\in \mathbb {N}$ with $a_{n}\geq {\tfrac {3}{2}}b|S|$ for all $n\geq N_{2}$. Hence, for all $n\geq N=\max\{N_{1},N_{2}\}$, there is: ${\tfrac {a_{n}}{b_{n}}}\geq {\tfrac {{\tfrac {3}{2}}b|S|}{{\tfrac {3}{2}}b}}=|S|\geq S$ So we have convergence $\lim _{n\to \infty }{\tfrac {a_{n}}{b_{n}}}=\infty$.

Case 2: $\lim _{n\to \infty }b_{n}=0$ with almost all $b_{n}$ being positive. Here, we assert $\lim _{n\to \infty }{\tfrac {a_{n}}{b_{n}}}={\frac {\infty }{+0}}=\infty$. A mathematical proof requires showing $\forall S\in \mathbb {R} \ \exists N\in \mathbb {N} \ \forall n\geq N:{\tfrac {a_{n}}{b_{n}}}\geq S$ Let $S\in \mathbb {R}$ 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 $b_{n}\leq {\tfrac {1}{2}}$ for all $n\geq N_{1}$. Since $\lim _{n\to \infty }a_{n}=\infty$ there is also an $N_{2}\in \mathbb {N}$ with $a_{n}\geq {\tfrac {1}{2}}|S|$ for all $n\geq N_{2}$. So for all $n\geq N=\max\{N_{1},N_{2}\}$, there is: ${\tfrac {a_{n}}{b_{n}}}\geq {\tfrac {{\tfrac {1}{2}}|S|}{\tfrac {1}{2}}}=|S|\geq S$ And again, we have convergence $\lim _{n\to \infty }{\tfrac {a_{n}}{b_{n}}}=\infty$.

All 4 cases are concluded in a theorem

Theorem (Quotient rule 1 for improperly convergent sequences)

Let $(a_{n})_{n\in \mathbb {N} }$ and $(b_{n})_{n\in \mathbb {N} }$ be real sequences with $\lim _{n\to \infty }a_{n}=\infty \in \mathbb {R}$.

• If $\lim _{n\to \infty }b_{n}=b>0$ or $\lim _{n\to \infty }b_{n}=0$ with almost all $b_{n}>0$, then $\lim _{n\to \infty }{\tfrac {a_{n}}{b_{n}}}=\infty$.
• If $\lim _{n\to \infty }b_{n}=b<0$ or $\lim _{n\to \infty }b_{n}=0$ with almost all $b_{n}<0$, then $\lim _{n\to \infty }{\tfrac {a_{n}}{b_{n}}}=-\infty$.

Direct comparison

Intuitively, if $(x_{n})$ is given and some "smaller" sequence $(y_{n})$ diverges to $\infty$, then also the "bigger" $(x_{n})$ must tend to $\infty$. This should still hold true if "$(x_{n})$ is bigger than $(y_{n})$" almost everywhere. Mathematically, we need to show $\forall S\in \mathbb {R} \ \exists N\in \mathbb {N} \ \forall n\geq N:x_{n}\geq S$ So let $S\in \mathbb {R}$ be given. Since $\lim _{n\to \infty }y_{n}=\infty$ there is an ${\tilde {N}}\in \mathbb {N}$ with $y_{n}\geq S$ for all $n\geq {\tilde {N}}$. Since $x_{n}\geq y_{n}$ for all but finitely many $n\in \mathbb {N}$ there is an $N\geq {\tilde {N}}$ with $x_{n}\geq y_{n}\geq S$ for all $n\geq N$. So indeed, $\lim _{n\to \infty }x_{n}=\infty$.

We conclude this in a theorem:

Theorem (Direct comparison for sequences)

Let $(x_{n})_{n\in \mathbb {N} }$ be a real sequence and $(y_{n})_{n\in \mathbb {N} }$ be another sequence with $x_{n}\geq y_{n}$ for almost all $n\in \mathbb {N}$ and let $\lim _{n\to \infty }y_{n}=\infty$. Then, $\lim _{n\to \infty }x_{n}=\infty$.

Example (Direct comparison for sequences)

Take the sequence $(x_{n})_{n\in \mathbb {N} }=\left({\tfrac {n^{n}}{n!}}\right)_{n\in \mathbb {N} }$. For $n\geq 2$ there is $x_{n}={\tfrac {n^{n}}{n!}}={\tfrac {n\cdot n\cdot \ldots \cdot n}{1\cdot 2\cdot \ldots \cdot n}}={\tfrac {n}{1}}\underbrace {\tfrac {n\cdot \ldots \cdot n}{2\cdot \ldots \cdot n}} _{\geq 1}\geq n=y_{n}.$ In addition $\lim _{n\to \infty }y_{n}=\lim _{n\to \infty }n=\infty$. So by direct comparison, $\lim _{n\to \infty }{\tfrac {n^{n}}{n!}}=\infty$

Of course, a similar statement holds true for $x_{n}\leq y_{n}$ and $\lim _{n\to \infty }y_{n}=-\infty$. Then also $\lim _{n\to \infty }x_{n}=\infty$ . This can easily seen by considering the sequences $(|x_{n}|)_{n\in \mathbb {N} }$ and $(|y_{n}|)_{n\in \mathbb {N} }$.