a) Rather prove it differently, your proof seems a little odd.
You know that and converges, so converges.
Since you know that . Thus and by the comparison test the first series must converge.
b) Is perfectly fine.
PROBLEM STATEMENT:
(a)Suppose that { } and { } are sequences of nonnegative numbers and let . Show that converges iff both and converge. Hint: .
(b)Prove that if converges, where , then converges, too. Hint: We already know that converges.
MY WORK:
(a) We use the comparison test here which says that if for , where is some fixed integer, and if converges, then converges.
Supposing that converges, then the larger of and converges. Suppose . Since converges, and , converges. Now by the comparison test, converges. The same is true if .
Now, supposing that both and converge, since and , must converge by the comparison test.
(b)Here, we use the Schwarz inequality which states that if and are complex numbers, then .
From here, we directly get . and since both converges, and converges, then converges.