That is NOT correct. For example, neither nor converge but their sum is converges trivially.
If diverges to positive infinity and , then diverges to infinity, by the comparison test, whether converges or not.
In fact, I would have handled the initial problem by noting that, since and so diverges by the comparison test.
The problem is equivalent to showing that if converges, then converges.
Or, that if converges, then converges. (Applying this result to gives the previous one)
So, assume that converges.
Note that we have , hence there are (that could be explicitated, but it's no use) such that, for all ,
However, since converges, is bounded: for some and all , hence converges absolutely.
Moreover, it is a "classical" fact that converges (cf. below).
As a consequence, we have
where . Each of the series converges, except (perhaps) the first one. Therefore, the first one converges as well.
.
This concludes.
Appendix: if converges, then converges. This is a consequence of "Abel summation". Let , so that converges to and is therefore bounded: for all and some . We have, for all ,
,
and the term on the right end converges to a finite limit as since (general term of a convergent series), and .