## Proof on Sequences: Sum of a convergent and divergent diverges

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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.
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 Mentor 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.
 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

Mentor

## Proof on Sequences: Sum of a convergent and divergent diverges

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