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. $\left|a_n+b_n-L-a_n+M\right|\le\left|a_n+b_n-L\right|+\left|-a_n+M\right|<\varepsilon$