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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!

Marko Karbevski
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    Well you could try: $a_n = n^2$, $b_n = 1/n$ – Winther Feb 07 '15 at 23:39
  • @Winther but then the limit exists and is infinity :/ – Marko Karbevski Feb 07 '15 at 23:40
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    Do you accept $\infty$ as a limit? – ploosu2 Feb 07 '15 at 23:40
  • Well, might as well take $a_n=n,b_n=n$, @Winther. I think OP is allowing the limit to be $\infty$. – Thomas Andrews Feb 07 '15 at 23:41
  • @ploosu2 Yes i do. – Marko Karbevski Feb 07 '15 at 23:41
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    I'm always amused by the misuse of the word "monotonous" for "monotonic." "Monotonous" means "boring/uninteresting" in English. – Thomas Andrews Feb 07 '15 at 23:50
  • Glad to be of amusement ;D. My answer was too hastily thinken. – ploosu2 Feb 07 '15 at 23:51
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    @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: http://math.stackexchange.com/questions/365717/is-monotonous-ever-used-as-a-synonym-for-monotonic-in-math – Thomas Andrews Feb 07 '15 at 23:57

2 Answers2

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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:

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

Thomas Andrews
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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.

clark
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