Is my proof correct?

  • Thread starter bavman
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  • #1
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Homework Statement



{an} and {bn} are convergent,

if lim an (n->inf) = M and lim bn (n->inf) = L

Prove that lim an -bn (n->inf) = M - L

Using Epsilon/n proof

Homework Equations





The Attempt at a Solution



Heres my attempt at it:

Given e > 0 we want to find N s.t. for all n>N

|(an - bn) - (M-L)| < e

|(an - bn) - (M-L)| <= |an-M|-|bn-L|

there exist N1 (element of natural number) s.t. for all n>N, |an - M| < e/2

there exist N2 (element of natural number) s.t. for all n>N, |bn - L| < e/2

let N = max{N1, N2} for all n>N

|(an - bn) - (M-L)| <= |an-M|-|bn-L|<e/2 + e/2 < e
 

Answers and Replies

  • #2
You applied the triangle inequality incorrectly.

You should have written

[tex]|a_n - M| + |- ( b_n-L)|[/tex].
If you have a minus sign between the absolute values then you will run into problems since the difference maybe negative. You want to have an inequality like

0< something < e.


That ensures that you can make "something" as small as you like since it is sandwiched between zero and epsilon.
Apart from that, the logic is sound.
 

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