# Thread:

1. ## Show that the sum of a convergent and divergent series is divergent

I am completely stumped on this one, so I could really use a hand with it.

Show that if the series $\sum a_k$ converges and the series $\sum b_k$ diverges, then the series $\sum (a_k+b_k)$ diverges.
A rigorous proof is expected, and we are to prove this by contradiction.

I don't know how I could answer this. This is all I really know that could possibly help me answer it.
If $\sum_{k=0}^{\infty}a_k$ converges and $\sum_{k=0}^{\infty}b_k$ converges, then $\sum_{k=0}^{\infty}(a_k+b_k)$
Moreover, if $\sum_{k=0}^{\infty}a_k=L$ and $\sum_{k=0}^{\infty}b_k=M$, then $\sum_{k=0}^{\infty}(a_k+b_k)=L+M$

I could really use some help with this one. I wasn't able to see my Calculus professor concerning the matter due to an unexplained absence on his part.

2. Hint: Argue by contradiction.
Start by assuming that the series $\sum (a_k+b_k)$ converges.

3. Originally Posted by General
Hint: Argue by contradiction.
Start by assuming that the series $\sum (a_k+b_k)$ converges.
Alright, I've tried your advice out and here's the work I currently have.

Assume $\sum (a_k+b_k)$ converges.
Since $\sum (-1)a_k=-\sum a_k$ converges,
$-\sum a_k + \sum (a_k+b_k) = \sum b_k$
This implies that $\sum b_k$ converges. This is a contradiction, since $\sum b_k$ diverges.

Is this proof correct?

4. Very good!
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