Question

Prove that if the nonnegative series \(\sum_{n=1}^{\infty} a_{n}\) and \(\sum_{n=1}^{\infty} b_{n}\) converge, then so does the series \(\sum_{n=1}^{\infty} a_{n} b_{n}\).

Short answer

The product of two convergent nonnegative series is also a convergent series.

Step-by-step solution

Preliminary Analysis

We know that \(a_n\) and \(b_n\) are nonnegative and converge. They must therefore be bounded, i.e., there exists \(M \geq 0\) such that \(0 \leq a_n, b_n \leq M\) for every \(n \geq 1\). Also, because \(a_n\) and \(b_n\) are nonnegative and converge, their series \(\sum_{n=1}^{\infty} a_n\) and \(\sum_{n=1}^{\infty} b_n\) also converge, which means the sequence of their partial sums is bounded.

Setting up the proof by contradiction

Assume, on the contrary, that the series \(\sum_{n=1}^{\infty} a_n b_n\) diverges. This means that the sequence of partial sums of the series is unbounded, i.e., there exists a subsequence of partial sums that is strictly increasing and goes to +∞.

Finding the contradiction

But each term in the series \(\sum_{n=1}^{\infty} a_n b_n\) is less than or equal to \(M^2\). So the sequence of partial sums of this series is increasing and bounded above by \(N*M^2\), which leads to a contradiction as a bounded sequence can't go to infinity.

Conclusion

Therefore, the assumption that the series \(\sum_{n=1}^{\infty} a_n b_n\) diverges is false, and thus, the series \(\sum_{n=1}^{\infty} a_n b_n\) converges.