Proof on Sequences: Sum of a convergent and divergent diverges

[Heute]

Homework Statement

Prove if sequence a_{n} converges and sequence b_{n} diverges, then the sequence a_{n}+b_{n} also diverges.

Homework Equations

The Attempt at a Solution

My professor recommended a proof by contradiction. That is, suppose a_{n}+b_{n} does converge. Then, for every ε > 0, there exists a natural number N_{1} so that n > N_{1} implies |a_{n}+b_{n} - L|< ε

We already know there exists N_{2} so that n > N_{2} implies |a_{n} - M| < ε. So let N = max{N_{1}, N_{2}}. Then n > N means we know a_{n} is "very close" to M. My purpose in this is to try and show that this implies b_{n} has a limit (that is, it converges) providing a contradiction. However, I'm not sure how to go about this.

 

[SammyS]

That's the general idea !

Rewrite \left|a_n+b_n-L\right| as \left|a_n-M+b_n-(L-M)\right|\,, then use the triangle inequality.

 

[Heute]

Ah! I think I've got it.

Proof: Assume a_{n} is convergent and b_{n} is divergent.

Now suppose that a_{n}+b_{n} is convergent.
Then [for every ε > 0 there exists a natural number N_{1} so that
n > N_{1} implies |a_{n}+b_{n} - L|< ε/2

We know by our assumption that there also exists natural number N_{2} so that
n > N_{2} implies |a_{n}-M| < ε/2

Now let N = max{N_{1},N_{2}}
Then n > N implies
|a_{n}+b_{n} - L|< ε/2 and
|a_{n}-M| < ε/2
So |a_{n}+b_{n} - L|+ |-(a_{n}-M)| < ε
and |a_{n}+b_{n} - a_{n} - (L-M)| < ε
(by the triangle inequality)
But a_{n}+b_{n} - a_{n} = b_{n}
so this last statement implies |b_{n} - (L-M)| < ε
which implies b_{n} converges, which is a contradiction
Therefore, a_{n}+b_{n} diverges

 

[SammyS]

Make it clear that by the triangle ineq. \left|a_n+b_n-L-a_n+M\right|\le\left|a_n+b_n-L\right|+\left|-a_n+M\right|&lt;\varepsilon