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Product rule of convergence sequence proof

Post Icon Posted: Submitted by tsl42 on 21 October 2007 - 7:53pm.

Joined: 2007-10-21
Posts: 1

Anybody know the proof

‹ Analysis III Too much work ›
Post Icon Posted: 21 October 2007 - 11:13pm

Joined: 2006-08-31
Posts: 696

That is:

given
$ a_n \longrightarrow a $ as $ n\rightarrow\infty $
and
$ b_n \longrightarrow b $ as $ n\rightarrow\infty $

prove:
$ a_nb_n \longrightarrow ab $ as $ n\rightarrow\infty $
Post Icon Posted: 22 October 2007 - 12:22am

Joined: 2006-10-09
Posts: 327
Obviously there's no giving out of answers to assignment sheets on here, but what I'd suggest is to look at the equation you're given:
$$(a_n-a)(b_n-b)+b(a_n-a)+a(b_n-b)=(a_nb_n-ab)$$
and consider, based on the fact you're asked about product and sum rules for null sequences (i.e. if $ a_n\rightarrow 0 $ and $ b_n\rightarrow 0 $ then $ a_nb_n\rightarrow 0 $ and $ \gamma a_n + \delta b_n \rightarrow 0 $) see if you can identify what all the parts are...

Also remember that $ a_n\rightarrow a $ implies $ (a_n-a)\rightarrow 0 $.

Hope that was of help!

Jamie

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