Sum of convergent and divergent series

[tarheelborn]

Homework Statement

Prove that if $\displaystyle\sum_{n=1}^\infty a_n$ converges and $\displaystyle\sum_{n=1}^\infty b_n$ diverges, then $\displaystyle\sum_{n=1}^\infty (a_n+b_n)$ diverges.

Homework Equations

I know that the limit of {a_n} = 0 because it is convergent, but I can't say anything about how {b_n} diverges.

The Attempt at a Solution

I know that I need to show that sum of the partial sums of the two series diverge, but I am not sure how to show it.

 

[jbunniii]

Try a proof by contradiction. If

\sum_{n=1}^\infty (a_n + b_n)

converges, and so does

\sum_{n=1}^\infty a_n

then what does that imply about

\sum_{n=1}^\infty b_n

 

[tarheelborn]

OK, so something like
Suppose (a_n+b_n) converges. Then, by the converse of the theorem that says if two infinite series are convergent, then their sum is convergent, a_n converges and b_n converges. But this is a contradiction since, by hypothesis, b_n diverges. Hence (a_n+b_n) must diverge.

Will this work? I am trying to improve my proof technique; I have had a bad habit of assuming too much. Thanks so much for your help!

 

[jbunniii]

OK, so something like
Suppose (a_n+b_n) converges. Then, by the converse of the theorem that says if two infinite series are convergent, then their sum is convergent, a_n converges and b_n converges. But this is a contradiction since, by hypothesis, b_n diverges. Hence (a_n+b_n) must diverge.

Will this work? I am trying to improve my proof technique; I have had a bad habit of assuming too much. Thanks so much for your help!

It's not "by the converse of the theorem that says if two infinite series are convergent, then their sum is convergent". It is by that theorem itself. Why is that? It is because

b_n = (a_n + b_n) + (-a_n)

You are given that

\sum a_n

converges. Therefore

\sum(-a_n)

also converges.

Now, what would happen if

\sum(a_n + b_n)

were convergent?

 

[tarheelborn]

Thanks!