Prove Sum of Converging Sequences = L+M, Counter Example

[Design]

Homework Statement

Prove that if two sequences a_n and bⁿ converge to L and M respectively, then the sum of the sequences converge to L+M.
Present a counter example to show the sum of two divergent sequences need not be divergent.

The Attempt at a Solution

We have a_n -> L and b_n -> M we want a_n + b_n -> L+M.

So I have |a_n - L | < Ε for every n > N
and |b_n - L | Ε every n > N

What is my next step?

thank you

 

[╔(σ_σ)╝]

Use the triangle inequality and an epsilon/2 argument.

 

[Design]

Is my starting point totally off?

 

[╔(σ_σ)╝]

No, it is quite good. But you may want to modify what you have in order to distinguise the n's for which a_n and b_n can be made smaller than epsilon. It would also be useful to put a_n and b_n in the epsilon/2 neighbourhood of L and M. You will see why later.

|a_n+ b_n -(L+M)|=|a_n-L + b_n -M|

Use the triangle inequality on the above and then select your n_0 and then use the modified part that you already have.

 

[Design]

How can i make an and bn with the E/2 argument, don't understand that part. I know we are trying to prove that statement.

 

[radou]

As ╔(σ_σ)╝ noted, it's only a point of rearranging terms, making use of the triangle inequality and chosing some ε > 0, and some ε/2 > 0. I.e. if an and bn coverge, then for ε/2 (whatever ε is) there exist some positive integers N1 and N2 such that |an - L| < ε/2 whenever N >= N1 and |bn - M| < ε/2 whenever N >= N2. For which integers will both inequalities be satisfied?

 

[Design]

maximum of n1 n2?

 

[╔(σ_σ)╝]

Yes :-).

Show us what you have done.