# Divergence to infinity: rules – Serlo

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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!

Exercise

Find examples for sequences ${\displaystyle (a_{n})}$ and ${\displaystyle (b_{n})}$ as above, such that

1. ${\displaystyle \lim _{n\to \infty }a_{n}b_{n}=0}$
2. ${\displaystyle \lim _{n\to \infty }a_{n}b_{n}=c}$ where ${\displaystyle c\neq 0}$
3. ${\displaystyle \lim _{n\to \infty }a_{n}b_{n}=\infty }$
4. ${\displaystyle \lim _{n\to \infty }a_{n}b_{n}=-\infty }$
5. ${\displaystyle (a_{n}b_{n})}$ is bounded and diverges
6. ${\displaystyle (a_{n}b_{n})}$ is unbounded, but does not converge improperly

Solution

Part 1: For instance, with ${\displaystyle a_{n}=n}$ and ${\displaystyle b_{n}={\tfrac {1}{n^{2}}}}$, there is ${\displaystyle \lim _{n\to \infty }a_{n}=\infty }$, ${\displaystyle \lim _{n\to \infty }b_{n}=0}$ and ${\displaystyle \lim _{n\to \infty }a_{n}b_{n}=\lim _{n\to \infty }{\tfrac {1}{n}}=0}$

Part 2: For instance, with ${\displaystyle a_{n}=n}$ and ${\displaystyle b_{n}={\tfrac {c}{n}}}$, there is ${\displaystyle \lim _{n\to \infty }a_{n}=\infty }$, ${\displaystyle \lim _{n\to \infty }b_{n}=0}$ and ${\displaystyle \lim _{n\to \infty }a_{n}b_{n}=\lim _{n\to \infty }c=c}$

Part 3: For instance, with ${\displaystyle a_{n}=n^{2}}$ and ${\displaystyle b_{n}={\tfrac {1}{n}}}$, there is ${\displaystyle \lim _{n\to \infty }a_{n}=\infty }$, ${\displaystyle \lim _{n\to \infty }b_{n}=0}$ and ${\displaystyle \lim _{n\to \infty }a_{n}b_{n}=\lim _{n\to \infty }n=\infty }$

Part 4: For instance, with ${\displaystyle a_{n}=n^{2}}$ and ${\displaystyle b_{n}=-{\tfrac {1}{n}}}$, there is ${\displaystyle \lim _{n\to \infty }a_{n}=\infty }$, ${\displaystyle \lim _{n\to \infty }b_{n}=0}$ and ${\displaystyle \lim _{n\to \infty }a_{n}b_{n}=\lim _{n\to \infty }-n=-\infty }$

Part 5: For instance, with ${\displaystyle a_{n}=n}$ and ${\displaystyle b_{n}={\tfrac {(-1)^{n}}{n}}}$, there is ${\displaystyle \lim _{n\to \infty }a_{n}=\infty }$, ${\displaystyle \lim _{n\to \infty }b_{n}=0}$ and ${\displaystyle (a_{n}b_{n})=((-1)^{n})}$ is bounded and diverges

Part 6: For instance, with ${\displaystyle a_{n}=n^{2}}$ and ${\displaystyle b_{n}={\tfrac {(-1)^{n}}{n}}}$, there is ${\displaystyle \lim _{n\to \infty }a_{n}=\infty }$, ${\displaystyle \lim _{n\to \infty }b_{n}=0}$ and ${\displaystyle (a_{n}b_{n})=((-1)^{n}n)}$ is unbounded, but does not converge improperly

## 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 ${\displaystyle (a_{n})}$ 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, ${\displaystyle \lim _{n\to \infty }a_{n}b_{n}=-\infty }$.

Exercise

Prove this.

Solution

We need to show:

${\displaystyle \forall S\in \mathbb {R} \ \exists n\in \mathbb {N} \ \forall n\geq N:a_{n}b_{n}\leq S}$

So let ${\displaystyle S\in \mathbb {R} }$ be given. Since ${\displaystyle \lim _{n\to \infty }b_{n}=b<0}$ , we have that for any ${\displaystyle \epsilon >0}$ some ${\displaystyle N_{1}\in \mathbb {N} }$ exists with ${\displaystyle |b_{n}-b|<\epsilon }$ for all ${\displaystyle n\geq N_{1}}$. Now, set ${\displaystyle \epsilon =-{\tfrac {b}{2}}}$. Then ${\displaystyle \forall n\geq N_{1}}$: ${\displaystyle |b_{n}-b|<-{\tfrac {b}{2}}}$, which especially includes ${\displaystyle b_{n}<{\tfrac {b}{2}}<0}$. Since ${\displaystyle \lim _{n\to \infty }a_{n}=\infty }$ there is also an ${\displaystyle N_{2}\in \mathbb {N} }$ with ${\displaystyle a_{n}\geq -{\tfrac {2|S|}{b}}\geq 0}$ for all ${\displaystyle n\geq N_{1}}$. Now, for ${\displaystyle n\geq N=\max\{N_{1},N_{2}\}}$ there is

${\displaystyle a_{n}b_{n}\leq -{\tfrac {2|S|}{b}}{\tfrac {b}{2}}=|S|\leq S}$

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

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 ${\displaystyle {\overline {\mathbb {R} }}=\mathbb {R} \cup \{\pm \infty \}}$.

Theorem (Product rule for improperly convergent sequences)

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

${\displaystyle \lim _{n\to \infty }a_{n}b_{n}={\begin{cases}\infty &{\text{ if }}b>0{\text{ or }}b=\infty ,\\-\infty &{\text{ if }}b<0{\text{ or }}b=-\infty .\end{cases}}}$

### Sum rule

Let again ${\displaystyle (a_{n})}$ 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:

Theorem (Sum rule for improperly convergent sequences)

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

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

### Inversion

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

Exercise

Prove that in this case, ${\displaystyle \left({\tfrac {1}{a_{n}}}\right)}$ again converges to 0.

Solution

Here, we also 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 ${\displaystyle a_{n}\leq -{\tfrac {1}{\epsilon }}-1<0}$. Hence, ${\displaystyle |a_{n}|\geq \left|-{\tfrac {1}{\epsilon }}-1\right|={\tfrac {1}{\epsilon }}+1}$. So ${\displaystyle \forall n\geq N}$ there is

${\displaystyle \left|{\tfrac {1}{a_{n}}}\right|\leq {\tfrac {1}{{\tfrac {1}{\epsilon }}+1}}<{\tfrac {1}{\tfrac {1}{\epsilon }}}=\epsilon }$

${\displaystyle \lim _{n\to \infty }a_{n}=0}$.

We conclude these findings in a theorem:

Theorem (Inversion rule 1 for improperly convergent series)

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

${\displaystyle \lim _{n\to \infty }{\tfrac {1}{a_{n}}}=0}$

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

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

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.

Exercise

Prove, that in this case ${\displaystyle \lim _{n\to \infty }a_{n}={\frac {1}{-0}}=-\infty }$.

Solution

This is a kind of "intelligent copycat exercise": We take the proof above and perform some minor changes in the presigns. Our aim is to show:

${\displaystyle \forall S\in \mathbb {R} \ \exists N\in \mathbb {N} \ \forall n\geq N:a_{n}\leq S}$

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 negative, 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 ${\displaystyle -a_{n}\geq |S|}$, and hence ${\displaystyle a_{n}\leq -|S|\leq S}$ for all ${\displaystyle n\geq N}$. So we get ${\displaystyle \lim _{n\to \infty }a_{n}=-\infty }$.

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

Theorem (Inversion rule 2 for improperly convergent series)

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

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

Example (Inversion rule)

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

${\displaystyle \lim _{n\to \infty }{\tfrac {1}{\tfrac {n^{k}}{z^{n}}}}=\lim _{n\to \infty }{\tfrac {z^{n}}{n^{k}}}=\infty }$

Analogously, for ${\displaystyle z>1}$:

${\displaystyle \lim _{n\to \infty }{\tfrac {n!}{z^{n}}}=\infty }$

### Quotient rule

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|

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

Theorem (Quotient rule 1 for improperly convergent sequences)

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

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 ${\displaystyle \lim _{n\to \infty }{\tfrac {a_{n}}{b_{n}}}=\infty }$.

Exercise

Prove that for ${\displaystyle \lim _{n\to \infty }b_{n}=b<0}$ and for ${\displaystyle \lim _{n\to \infty }b_{n}=0}$, with almost all ${\displaystyle b_{n}}$ being negative, there is: ${\displaystyle \lim _{n\to \infty }{\tfrac {a_{n}}{b_{n}}}=-\infty }$

Solution

Case 1: ${\displaystyle \lim _{n\to \infty }b_{n}=b<0}$. We need to prove:

${\displaystyle \forall S\in \mathbb {R} \ \exists N\in \mathbb {N} \ \forall n\geq N:{\tfrac {a_{n}}{b_{n}}}\leq S}$

Let ${\displaystyle S\in \mathbb {R} }$ be given. Since ${\displaystyle \left(b_{n}\right)}$ converges to ${\displaystyle b<0}$ , there is an ${\displaystyle N_{1}\in \mathbb {N} }$, such that ${\displaystyle b_{n}\geq {\tfrac {3}{2}}b}$ for all ${\displaystyle n\geq N_{1}}$. Since ${\displaystyle \lim _{n\to \infty }a_{n}=-\infty }$ there is an ${\displaystyle N_{2}\in \mathbb {N} }$ with ${\displaystyle a_{n}\leq -{\tfrac {3}{2}}b|S|}$ for all ${\displaystyle n\geq N_{2}}$. Hence, for all ${\displaystyle n\geq N=\max\{N_{1},N_{2}\}}$ there is:

${\displaystyle {\tfrac {a_{n}}{b_{n}}}\leq -{\tfrac {{\tfrac {3}{2}}b|S|}{{\tfrac {3}{2}}b}}=-|S|\leq S}$

So we have convergence ${\displaystyle \lim _{n\to \infty }{\tfrac {a_{n}}{b_{n}}}=-\infty }$.

Case 2: ${\displaystyle \lim _{n\to \infty }b_{n}=0}$ and almost all ${\displaystyle b_{n}}$ are negative. Again, we need to show:

${\displaystyle \forall S\in \mathbb {R} \ \exists N\in \mathbb {N} \ \forall n\geq N:{\tfrac {a_{n}}{b_{n}}}\leq S}$

Let ${\displaystyle S\in \mathbb {R} }$ be given. Since ${\displaystyle \left(b_{n}\right)}$ converges to ${\displaystyle 0}$ and almost all elements are negative, there is an ${\displaystyle N_{1}\in \mathbb {N} }$ with ${\displaystyle b_{n}\geq -{\tfrac {1}{2}}}$ for all ${\displaystyle n\geq N_{1}}$. Since ${\displaystyle \lim _{n\to \infty }a_{n}=-\infty }$ there is an ${\displaystyle N_{2}\in \mathbb {N} }$ with ${\displaystyle a_{n}\leq {\tfrac {1}{2}}|S|}$ for all ${\displaystyle n\geq N_{2}}$. Hence, for all ${\displaystyle n\geq N=\max\{N_{1},N_{2}\}}$there is:

${\displaystyle {\tfrac {a_{n}}{b_{n}}}\leq {\tfrac {{\tfrac {1}{2}}|S|}{-{\tfrac {1}{2}}}}=-|S|\leq S}$

And again, we have convergence ${\displaystyle \lim _{n\to \infty }{\tfrac {a_{n}}{b_{n}}}=-\infty }$.

All 4 cases are concluded in a theorem

Theorem (Quotient rule 1 for improperly convergent sequences)

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

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

### Direct comparison

Intuitively, if ${\displaystyle (x_{n})}$ is given and some "smaller" sequence ${\displaystyle (y_{n})}$ diverges to ${\displaystyle \infty }$, then also the "bigger" ${\displaystyle (x_{n})}$ must tend to ${\displaystyle \infty }$. This should still hold true if "${\displaystyle (x_{n})}$ is bigger than ${\displaystyle (y_{n})}$" almost everywhere. Mathematically, we need to show

${\displaystyle \forall S\in \mathbb {R} \ \exists N\in \mathbb {N} \ \forall n\geq N:x_{n}\geq S}$

So let ${\displaystyle S\in \mathbb {R} }$ be given. Since ${\displaystyle \lim _{n\to \infty }y_{n}=\infty }$ there is an ${\displaystyle {\tilde {N}}\in \mathbb {N} }$ with ${\displaystyle y_{n}\geq S}$ for all ${\displaystyle n\geq {\tilde {N}}}$. Since ${\displaystyle x_{n}\geq y_{n}}$ for all but finitely many ${\displaystyle n\in \mathbb {N} }$ there is an ${\displaystyle N\geq {\tilde {N}}}$ with ${\displaystyle x_{n}\geq y_{n}\geq S}$ for all ${\displaystyle n\geq N}$. So indeed, ${\displaystyle \lim _{n\to \infty }x_{n}=\infty }$.

We conclude this in a theorem:

Theorem (Direct comparison for sequences)

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

Example (Direct comparison for sequences)

Take the sequence ${\displaystyle (x_{n})_{n\in \mathbb {N} }=\left({\tfrac {n^{n}}{n!}}\right)_{n\in \mathbb {N} }}$. For ${\displaystyle n\geq 2}$ there is

${\displaystyle x_{n}={\tfrac {n^{n}}{n!}}={\tfrac {n\cdot n\cdot \ldots \cdot n}{1\cdot 2\cdot \ldots \cdot n}}={\tfrac {n}{1}}\underbrace {\tfrac {n\cdot \ldots \cdot n}{2\cdot \ldots \cdot n}} _{\geq 1}\geq n=y_{n}.}$

In addition ${\displaystyle \lim _{n\to \infty }y_{n}=\lim _{n\to \infty }n=\infty }$. So by direct comparison,

${\displaystyle \lim _{n\to \infty }{\tfrac {n^{n}}{n!}}=\infty }$

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