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 :

*

*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?

*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!
 A: 

*

*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:

*

*The sequence has a limit that is not $\infty$

*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 $$b_n =1 + (n-1)\cdot \left|\cos\left(\frac{n\pi}{2}\right)\right|$$
