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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!

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    $\begingroup$ Did you ignore the condition that the sequences are increasing? $\endgroup$ Nov 19, 2021 at 9:52
  • $\begingroup$ @KaviRamaMurthy I forgot to edit sorry , the question only says that they are positive $\endgroup$
    – Adamrk
    Nov 19, 2021 at 9:54
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    $\begingroup$ "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 $n \in \mathbb N^+.$ $\endgroup$
    – emonHR
    Nov 19, 2021 at 9:58
  • $\begingroup$ @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? $\endgroup$
    – Adamrk
    Nov 19, 2021 at 10:03
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    $\begingroup$ 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 $\lim_{n\to\infty}a_{n}\ne \infty$. $\endgroup$
    – user649348
    Nov 19, 2021 at 11:51

1 Answer 1

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

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