Problem 4. Prove that if ∑∞n=1 an converges absolutely and {bn}n≥1 is a bounded sequence,then the series ∑∞n=1 anbn also converges. (Hint: Use the Cauchy criterion.)

To prove this, we will use the Cauchy criterion for series convergence, which states that a series ∑an converges if and only if for every ε > 0, there exists an N such that for all m > n > N, |am + am+1 + ... + an| < ε.

Given that ∑an converges absolutely, this means that ∑|an| also converges. Therefore, by the Cauchy criterion, for every ε > 0, there exists an N1 such that for all m > n > N1, |a_n| + |a_n+1| + ... + |a_m| < ε.

Now, since {bn} is a bounded sequence, there exists a number M such that |bn| ≤ M for all n.

We want to show that ∑anbn also converges. So, we need to show that for every ε > 0, there exists an N such that for all m > n > N, |anbn + an+1bn+1 + ... + ambm| < ε.

Consider the absolute value of the partial sum:

|anbn + an+1bn+1 + ... + ambm| ≤ |anbn| + |an+1bn+1| + ... + |ambm| (by the triangle inequality)

≤ |an|M + |an+1|M + ... + |am|M (since |bn| ≤ M)

= M(|an| + |an+1| + ... + |am|).

Now, we know that for every ε > 0, there exists an N1 such that for all m > n > N1, |an| + |an+1| + ... + |am| < ε/M. Therefore, for all m > n > N1, M(|an| + |an+1| + ... + |am|) < ε.

So, by the Cauchy criterion, ∑anbn also converges.