# Simple two sequences limit multiplication is infinite and $(a_n)$ ,$(b_n)$ are positive then the limit of $a_n$ or $b_n$ is infinite

let $$(a_n)$$ and $$(b_n)$$be sequences such that $$\lim\limits_{n\to\infty }a_n \cdot b_n= \infty$$. if almost all of the elements in $$(a_n)$$ and $$(b_n)$$ are positive then $$\lim\limits_{n\to\infty }a_n= \infty$$ or $$\lim\limits_{n\to\infty }b_n= \infty$$ .

the statement is not true and it can be approached by a counter example such that

$$b_n$$=$$\begin{cases} n , n_{odd}\\ 1 , n_{even} \end{cases}$$ and $$a_n$$ = $$\begin{cases} 1 , n_{even}\\ n, n_{odd} \end{cases}$$

for $$n_{even}$$ we get $$a_n \cdot b_n = n \cdot 1 = \infty$$ and for $$n_{odd}$$ $$a_n \cdot b_n = 1 \cdot n = \infty$$ and both of the sequences are positive so they are also positive almost for all the sequence.

to prove $$\lim\limits_{n\to\infty }a_n\not= \infty$$ then there must exist $$M$$ such that for every $$N$$ there is $$n>N$$ that fulfills the following $$a_n \leq M$$

then we can take $$M=3$$ and let $$N \in \Bbb N$$ then $$n=2N+1$$ that fulfills $$n=2N+1 > 2N >N$$ and this number is a natural odd number so $$a_n =1 <3=M$$

my questions are :

1. When I prove that the limit is not infinite is it enough to show that it is only not infinite for odd number like what I did?
2. Is there a different way to show this other than a counter example using piecewise functions? if yes what is the other way? or if using a counterexample I would like to see a way not using a piecewise function

Thank you!

• Did you ignore the condition that the sequences are increasing? Nov 19, 2021 at 9:52
• @KaviRamaMurthy I forgot to edit sorry , the question only says that they are positive Nov 19, 2021 at 9:54
• "When I prove that the limit is bounded is it enough to show that it is only bounded for odd number like what I did?" $\rightarrow$ No. Def: A sequence $\{x_n \}$ is said to be bounded if $\exists M > 0$ such that $|x_n| \le M$ for all $n \in \mathbb N^+.$ Nov 19, 2021 at 9:58
• @emonHR Thank you , then does that mean that my example is not bounded? because if I check for $n_{even}$ then I get that it is not bounded.. am I missing something? Nov 19, 2021 at 10:03
• A common definition of $\lim_{n\to \infty} a_{n}=\infty$ is that for every $N\in \mathbb{N}$ there exists a $M\in \mathbb{N}$ such that $a_{n}>N$ for every $n\geq M$. Negating this statement we get that there exists a $N\in \mathbb{N}$ such that for every $M\in \mathbb{N}$ we can find a $n\geq M$ with $a_{n}\leq N$. You have shown this for $a_{n}$ because if we choose $N=2$ and let $M\in \mathbb{N}$ be arbitrary then we can choose $n=2N$ so that $a_{n}=a_{2N}=1<2=N$. This shows that $\lim_{n\to\infty}a_{n}\ne \infty$.
– user649348
Nov 19, 2021 at 11:51

1. When I prove that the limit is not infinite is it enough to show that it is only not infinite for odd number like what I did?

Yes. If a sequence has a limit, then all its subsequences have the same limit.

This means that if there exists a single subsequence for which the limit is not $$\infty$$, then the limit of the sequence cannot be $$\infty$$ as well.

Note that this only proves that the limit of the sequence is not $$\infty$$. You still have two other options:

1. The sequence has a limit that is not $$\infty$$
2. The sequence does not have a limit.

Is there a different way to show this other than a counter example using piecewise functions? if yes what is the other way? or if using a counterexample I would like to see a way not using a piecewise function

Disproving a statement $$\forall x: P(x)$$ is equivalent to proving the statement $$\exists x: \neg P(x)$$. So really, any proof that disproves your statement will, in one way or another, prove that a counterexample exists.

If you don't want to use a "piecewise" function, you can always redefine $$a_n$$ to be $$a_n =1 + (n-1)\cdot \left|\sin\left(\frac{n\pi}{2}\right)\right|$$ and now there are no "piecewise" functions in the definition of $$a_n$$. Note that all the values of $$a_n$$ stayed the same though :). Similarly, you can define $$b_n =1 + (n-1)\cdot \left|\cos\left(\frac{n\pi}{2}\right)\right|$$