Divergence to infinity: rules – Serlo

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!

Exercise

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

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

Solution

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

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

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

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

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

Part 6: For instance, with $a_{n}=n^{2}$ and $b_{n}={\tfrac {(-1)^{n}}{n}}$, there is $\lim _{n\to \infty }a_{n}=\infty$, $\lim _{n\to \infty }b_{n}=0$ and $(a_{n}b_{n})=((-1)^{n}n)$ 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 $(a_{n})$ 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, $\lim _{n\to \infty }a_{n}b_{n}=-\infty$.

Exercise

Prove this.

Solution

We need to show:

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

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

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

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

Those four cases can also be concluded into one statement. We introduce a practical extension of the real numbers: To the set $\mathbb {R}$ , 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: $\lim _{n\to \infty }a_{n}=-\infty$.

Exercise

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

Solution

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

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

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

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.

Exercise

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

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:

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

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

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

Theorem (Inversion rule 2 for improperly convergent series)

Let $(a_{n})_{n\in \mathbb {N} }$ 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 $\lim _{n\to \infty }{\tfrac {a_{n}}{b_{n}}}=\infty$.

Exercise

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

Solution

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

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

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

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

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

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

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

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

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

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 $(|y_{n}|)_{n\in \mathbb {N} }$.