If we are given monotonic real sequences $(a_n)_{n\ge1}, (b_n)_{n\ge1}$ then does the limit $\lim_{n \to \infty} a_n b_n $ always exists (+-infinity is also considered as limit point)? The case that needs looking into is obviously the one when, for example $a_n \to \infty$ and $b_n \to 0$ .

I've been thinking a bit and i can't seem to find a counterexample (i am pretty sure that a limit does not always exist)

Thanks in advance!

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    $\begingroup$ Well you could try: $a_n = n^2$, $b_n = 1/n$ $\endgroup$
    – Winther
    Feb 7, 2015 at 23:39
  • $\begingroup$ @Winther but then the limit exists and is infinity :/ $\endgroup$ Feb 7, 2015 at 23:40
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    $\begingroup$ Do you accept $\infty$ as a limit? $\endgroup$
    – ploosu2
    Feb 7, 2015 at 23:40
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    $\begingroup$ I'm always amused by the misuse of the word "monotonous" for "monotonic." "Monotonous" means "boring/uninteresting" in English. $\endgroup$ Feb 7, 2015 at 23:50
  • 1
    $\begingroup$ @ploosu2 The original question used "monotonous," too, so you aren't alone. It apparently is common amongst Germans. From a question about this I asked a while back: "In German it's monoton and it is often hard for us non-native speakers to guess which ending a Latin or Greek word will have in English. Moreover, the German word monoton has both meanings: monotonous and monotonic." See: math.stackexchange.com/questions/365717/… $\endgroup$ Feb 7, 2015 at 23:57

2 Answers 2


Let $a_n=2^{-\lfloor \log_2 n\rfloor}$ and $b_n=n$. Then $\limsup_{n\to\infty} a_nb_n = 2$ and $\liminf_{n\to\infty} a_nb_n=1$.

We see this because $a_nb_n = 2^{\{\log_2 n\}}$ where $\{x\}=x-\lfloor x\rfloor$ is the fractional part of $x$.

If you want strictly monotonic:

$$a_n= \frac{1}{2^{\lfloor\log_2 n\rfloor} n}, b_n=n^2$$


Let us do this when $a_n$ and $b_n$ are positive sequences.

Claim: Let $c_n$ be a positive valued sequence then there are monotonic sequences $a_n,b_n$ such that $c_n=a_n b_n$.

WLOG assume that $a_n$ is decreasing and $b_n$ is increasing. We do this recursively. Assume we have defined the first $k$-terms, then what we need to do is to find $a_{k+1}$ and $b_{k+1}$ such that $c_{k+1}= a_{k+1}b_{k+1}$ and $a_{k+1}$ is the smaller than all $a_1,a_2,\cdots, a_k$ and similarly for $b_{k+1}$. But this is always possible since $c_k=xy$ is a parabola.

Now for your question take $c_n$ non convergent and use the claim.


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