Show that $\limsup_{n \to \infty}a_nb_n= \limsup_{n \to \infty}a_n \cdot \lim_{n \to \infty}b_n$

Asked Jan 25 '16 at 13:14

Active Jan 25 '16 at 13:31

Viewed 248 times

Let $\{a_n\}_{n \geq0}$ and $\{b_n\}_{n \geq0}$ be two real sequences and suppose that $\lim_{n \to \infty} b_n=b \geq 0$ exist. Show that $$\limsup_{n \to \infty}a_nb_n= \limsup_{n \to \infty}a_n \cdot \lim_{n \to \infty}b_n$$

I am stacked on this problem. Could anyone give me a hint?

edited Jan 25 '16 at 13:23

Jimmy R.

35,868

asked Jan 25 '16 at 13:14

Taj Mohamed Bandalandabad

1

If $lim_n a_n b_n=\infty$ and $lim_n b_n=0$ then $(lim sup_n a_n)(lim_n b_n)=(\infty)\cdot 0$ which is undefinable. E.g, $a_n=2^n$ and $b_n=1/n$. – DanielWainfleet Jan 25 '16 at 13:22
@user254665 So there is a mistake in the question? One argument more should be that ${a_n}_{n \geq 0}$ has to be bound? – Taj Mohamed Bandalandabad Jan 25 '16 at 13:26
1

Duplicate: http://math.stackexchange.com/q/113121/ – Did Jan 25 '16 at 13:27
$a_n$ bounded, or that $lim_n b_n>0$ . – DanielWainfleet Jan 25 '16 at 13:29
@user254665 I don't understand your first comment – Taj Mohamed Bandalandabad Jan 25 '16 at 13:34

Show that $\limsup_{n \to \infty}a_nb_n= \limsup_{n \to \infty}a_n \cdot \lim_{n \to \infty}b_n$

0 Answers0