Prove or disprove that if $\sum a_n$ and $\sum b_n$ are both divergent, then $\sum (a_n \pm b_n)$ necessarily diverges. Prove or disprove that if $\sum a_n$ and $\sum b_n$ are both divergent, then $\sum (a_n \pm b_n)$ necessarily diverges.
I decided to disprove this by offering the example:
Let $a_n = n$.
Let $b_n = -n$.
Then $\sum (a_n + b_n) = \sum 0 = 0$, which converges.
However, my question says $\pm$, and in the $-$ case, we have $\sum (a_n - b_n) = \sum 2n = \infty$, which diverges. So, does this still count as a valid counter example or not? I partly proved it wrong so I think it's still valid.
 A: If $\sum(a_n+b_n)$ and $\sum(a_n-b_n)$ both converge, then since :
$$a_n=\frac12\left((a_n+b_n)+(a_n-b_n)\right)\quad\textrm{and}\quad b_n=\frac12\left((a_n+b_n)-(a_n-b_n)\right)$$
we see that $\sum a_n$ and $\sum b_n$ both converge.
In other words, if $\sum a_n$ diverges or $\sum b_n$ diverges, then $\sum(a_n+b_n)$ diverges or $\sum(a_n-b_n)$ diverges.
A: My reading of the statement 
"$\sum (a_n \pm b_n)$ necessarily diverges"
is that it means that both of the sums
$\sum (a_n + b_n)$ and $\sum (a_n - b_n)$ diverge.
You have proved that this is not necessarily true; you have a
counterexample in which one of the sums converges.
The wording of the original question seems highly ambiguous to me, however.
The statement "$\sum (a_n \pm b_n)$ diverges"
might be intended to say that either
$\sum (a_n + b_n)$ diverges or $\sum (a_n - b_n)$ diverges.
It really comes down to what is the meaning of the statement "$\sum (a_n \pm b_n)$ diverges."
A: Let $\sum \frac{1}{2n+1}$=$1+1/3+1/5+1/7+1/9+....$ and let $-\sum \frac{1}{2n}$=$-1/2-1/4-1/6-1/8...$. Boths series diverge, but when you add them (term by term) together, you get $1-1/2+1/3-1/4+1/5-1/6...$ and this is the well known Alternating Harmonic series with an answer of $\ln 2$, so (conditionally) convergent. 
