Does Proving An/Bn -> L Ensure An Converges to BnL as n Approaches Infinity?

[Fisicks]

I am given that An/Bn -> L, as n goes to infinity, where An and Bn are sequences. I also know that An and Bn converge to zero and have positive terms.
Pick some E>0, and by definition of the limit there exists an N such |An/Bn - L| < E for all n>N.
Because Bn converges to zero, the sequence is bound by M.

The next step is when my question comes in because I have never really shown that a sequence converges to a sequence.

1/M|An-BnL| < 1/Bn|An-BnL|= |An/Bn - L|< E

Therefore take E to be E/M, and An converges to BnL.

Correct?

 

[HallsofIvy]

It doesn't make sense to say that "a sequence converges to a sequence". What you are showing is that two sequences have the same limit. But if you have already said that \{A_n\} and \{B_n\} converge to 0, It is trivial that \{B_nL\} also converges to 0 and so that \{A_n\} and \{B_nL\} converge to the same limit- 0.

 

[Fisicks]

I should probably let you know my aim is to show that Bn+1/Bn <1 if I know that An+1/An <1. Would my method allow me to replace An+1 by Bn+1L and An by BnL in limit as n goes to infinity?