Let $\lim\limits_{n\to\infty }a_n \cdot b_n = \infty$ , If $0<b_n < a_n$ almost for every $n$ then $\lim\limits_{n\to\infty }a_n = \infty$

This is the exact same question here , according to the question asked I know that the statement is true.

This is what I tried and got stuck:

according to the definition of infinite limits , A sequence $(C_n)$ tends to infinity if for every $M \in \Bbb R$ there exists an $N \in \Bbb N$ such that for every $n \geq N$ we get $C_n > M$.

so according to the given information $\lim\limits_{n\to\infty }a_n \cdot b_n = \infty$ there exists an $N \in \Bbb N$ such that for every $n > N$ we get $a_n \cdot b_n >M$

and also from the information $0<b_n < a_n$ , there exists a $K \in \Bbb N$ such that for every $n > K$ we get $ 0 < b_n < a_n$ , if we multiply by $a_n$ we get $0<b_n \cdot a_n < a^2_{n}$

let $B=max({K,N})$ then for every $n>B$ we get $M<a_n \cdot b_n<a_n^2$ therefore $M< a_n^2$

I got stuck here because I cannot get to $M < a_n$ from here but I feel like what I did is correct in some way as it only relies on definitions and given information

Thanks for any tips and help!


1 Answer 1


Trick: Start with $M^{2}$, as it is also a positive number, then proceed your argument you will have $M^{2}<a_{n}^{2}$ and so $M<a_{n}$.

  • $\begingroup$ Thank you , obviously this simple trick worked. Is there another way you would approach this question? or recommend a different way than what I did ? as it relies too much on definition and less actual analyzing $\endgroup$
    – Adamrk
    Nov 20, 2021 at 13:25
  • 1
    $\begingroup$ Actually you can stop at $\sqrt{M}<a_{n}$ and since $\sqrt{M}\rightarrow M$ is one-to-one and onto map on $(0,\infty)$, $\sqrt{M}$ represents every positive real as $M>0$ is varying, but then this may not sound as a precise mathematical reasoning though. $\endgroup$
    – user284331
    Nov 20, 2021 at 13:30
  • $\begingroup$ Thanks again , I still haven't studied one to one and onto map and more advanced stuff , as I am still at the basics but I will keep that in mind :) $\endgroup$
    – Adamrk
    Nov 20, 2021 at 13:31

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