Convergence of Series: Proof for Σ(anbn) with Convergent Σan and Bounded bn

Let M be a bound for the sequence bn, i.e. |bn| ≤ M for all n. Then we have:

|anbn| ≤ |an|M

Since Σan converges, we know that |an| → 0 as n → ∞. Therefore, for any ε > 0, there exists N such that |an| < ε/M for all n ≥ N.

Now, for any n ≥ N:

|anbn| ≤ |an|M < ε

Therefore, by the comparison test, Σ(anbn) converges absolutely, and hence converges.