3

I recently came across this question here and there is something I don't understand.

In the question we are considering a sequence $a_n$ with $\lim \sup_{n \to \infty}a_n \le \rho$ and we want to show that $\lim \sup_{n \to \infty} a_n^{{(n-m)}/{n}} \le \rho$. The asker tried to split up the $a_n^{{(n-m)}/{n}}$ as

$$ a_n^{{(n-m)}/{n}} = a_n \cdot a_n^{- {m \over n}}$$

but the answerer gives an example for which this does not work: if $a_n = \frac{1}{n^{\frac{n}{m}}}$ then $a_n^{- {m \over n}}$ diverges.

What I don't understand is why this example works. Because for sequences that converge (and $\limsup$ of a sequence always converges) we have

$$ \lim a_n b_n = \lim a_n \lim b_n$$

so it seems to me that splitting up a product should always be allowed. Yet, in this case it's not.

Please could someone explain to me what I'm missing and why the product rule for limits cannot be applied here?

Community
  • 1
a student
  • 4,365
  • Even though the equality is not true for limsup, you have at least inequality (for positive numbers): http://math.stackexchange.com/questions/197858/proof-of-lim-supa-nb-n-leq-lim-supa-n-limsupb-n – Martin Sleziak Oct 28 '15 at 06:57
  • And it is also useful to know that it would work if one of the sequences converges to a positive number, then the equality works: http://math.stackexchange.com/questions/768011/product-of-lim-sups – Martin Sleziak Oct 28 '15 at 07:01

2 Answers2

1

HINT: Think a bit about this case and see if the penny drops:

$$a_n=(-1)^n, \quad b_n=(-1)^{n+1}.$$

Aloizio Macedo
  • 34,292
  • Okay, say we let $A_n = \sup_{k \ge n} a_n$ and similarly $B_n = \sup_{k \ge n} b_n$. Then $A_n = B_n = -1 = A_n B_n$. So $$ \limsup a_n b_n \neq \limsup a_n \limsup b_n$$ So we see that the limit rule does not apply in this case but my question is why does it not. I still don't see it. – a student Oct 24 '15 at 03:57
  • 1
    Perhaps this might give some intuition: (similar to the answer above, but I find this marginally more intuitive) Take two sequences $a_n$ and $b_n$ that both oscillate between $0$ and $1$ but their oscillations are offset from each other. I.e. so that $a_{2n+1} = 0, a_{2n} = 1$, whereas $b_{2n+1} = 1, b_{2n} = 0$. Then note that $a_nb_n$ is always $0$. But the $\limsup$ picks up the highest value of the oscillation ($1$ in this case) for each of ${a_n}$ and ${b_n}$. In other words, ${a_nb_n}$ is insensitive to the oscillations of ${a_n}$ and ${b_n}$ because of this offset. –  Oct 24 '15 at 04:19
  • @SouparnaP Thank you, yes this is marginally more insightful. But I still struggle seeing the mistake in my thinking: Taking the lim sup of a sequence $a_n$ gives a new (convergent!) sequence $A_n = \sup{k \ge n} a_n$. Similarly for $b_n$. So we should be able to apply the limit theorem for the product to the new sequences $A_n, B_n$ but somehow this is wrong. – a student Oct 24 '15 at 04:23
  • Ok, I found the mistake. – a student Oct 24 '15 at 04:28
  • Well applying the limit theorem to $A_n$ and $B_n$ gives you $\lim A_nB_n = \lim A_n \lim B_n = \limsup a_n \limsup b_n$. But now, is $\lim A_nB_n = \limsup a_nb_n$? –  Oct 24 '15 at 04:28
  • @SouparnaP Yeah, exactly! :-D – a student Oct 24 '15 at 04:29
  • @astudent I think it is good that you spotted the mistake by yourself (this is why I gave you a hint saying "look at this special case" instead of a full explanation). I am of the opinion that in certain situations, being told what is the mistake is a palliative solution that creates only momentary comfort. Furthermore, it is worth mentioning that (not only on this case, but throughout mathematics) looking at very particular special cases can be very fruitful. A quote by Hilbert: "The art of doing mathematics consists in finding that special case which contains all the germs of generality." – Aloizio Macedo Oct 24 '15 at 04:48
1

With $a_n=1/(n^{n/m}$ and $b_n=a_n^{-m/n}=n$ we have $\lim a_n=0$ The expression $\lim b_n=\infty$ is symbolic short-hand and so is the sentence "$b_n$ goes to $\infty$" because there is no number $\infty$ for which all the usual rules of arithmetic apply. In particular "$\lim a_n .\lim b_n=0.\infty$" is meaningless. The product-limit rule is only valid when the limits are both real numbers.

DanielWainfleet
  • 57,985