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 ${\displaystyle (a_{n})}$ (improperly) converges to ${\displaystyle \infty }$ and a second sequence ${\displaystyle (b_{n})}$ (properly) converges to ${\displaystyle 0}$, one cannot make any statement about the convergence of their product!

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!

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

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

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

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

Case 4: ${\displaystyle \lim _{n\to \infty }b_{n}=b<0}$. Here, ${\displaystyle \lim _{n\to \infty }a_{n}b_{n}=-\infty }$.

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

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

• For ${\displaystyle a_{n}=n}$ and ${\displaystyle b_{n}=-n}$ there is ${\displaystyle \lim _{n\to \infty }a_{n}+b_{n}=\lim _{n\to \infty }0=0}$.
• For ${\displaystyle a_{n}=2n}$ and ${\displaystyle b_{n}=-n}$ there is ${\displaystyle \lim _{n\to \infty }a_{n}+b_{n}=\lim _{n\to \infty }n=\infty }$.

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

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

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

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

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

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

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

Both cases can be concluded in a theorem:

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

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

Case 2: ${\displaystyle \lim _{n\to \infty }a_{n}=-\infty }$.

We conclude these findings in a theorem:

The question is now: can we define a "converse of the inversion rule" which holds under more special assumptions? The sequence ${\displaystyle ((-1)^{n}n)}$ is not diverging to ${\displaystyle \infty }$ or ${\displaystyle -\infty }$ because it keeps changing presign, so there is a subsequence of it converging to ${\displaystyle \infty }$ and a subsequence converging to ${\displaystyle -\infty }$. We can avoid this by forbidding a change of presign in ${\displaystyle (a_{n})}$. 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 ${\displaystyle (a_{n})}$ be a sequence with ${\displaystyle \lim _{n\to \infty }{\tfrac {1}{a_{n}}}=0}$, all sequence elements being ${\displaystyle \neq 0}$ and all but finitely many sequence elements being positive. Then, intuitively ${\displaystyle \lim _{n\to \infty }a_{n}={\frac {1}{+0}}=\infty }$. For a mathematical proof, we need to show that ${\displaystyle \forall S\in \mathbb {R} \ \exists N\in \mathbb {N} \ \forall n\geq N:a_{n}\geq S}$ Let ${\displaystyle S\in \mathbb {R} }$ be given. Since ${\displaystyle \left({\tfrac {1}{a_{n}}}\right)}$ is a null sequence, for ${\displaystyle \epsilon ={\tfrac {1}{|S|}}>0}$ we can find an ${\displaystyle {\tilde {N}}\in \mathbb {N} }$ with ${\displaystyle \left|{\tfrac {1}{a_{n}}}\right|<{\tfrac {1}{|S|}}}$ for all ${\displaystyle n\geq {\tilde {N}}}$. Since almost all elements of ${\displaystyle (a_{n})}$ are positive, there is an ${\displaystyle N\geq {\tilde {N}}}$ with ${\displaystyle \left|{\tfrac {1}{a_{n}}}\right|={\tfrac {1}{a_{n}}}<{\tfrac {1}{|S|}}}$ for all ${\displaystyle n\geq N}$. Therefore ${\displaystyle a_{n}\geq |S|\geq S}$ for all ${\displaystyle n\geq N}$. So we get ${\displaystyle \lim _{n\to \infty }a_{n}=\infty }$.

Case 2: Let now ${\displaystyle (a_{n})}$ v ${\displaystyle \lim _{n\to \infty }{\tfrac {1}{a_{n}}}=0}$, all sequence elements being ${\displaystyle \neq 0}$ but this time, almost all of them are negative.

The converse of the inversion rule is also concluded into a theorem.

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

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

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

We conclude

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

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

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