Problem 115 - Essential Calculus

Question

Prove that if \(\sum\left|a_{n}\right|\) converges, then \(\sum a_{n}^{2}\) converges. Is the converse true? If not, give an example that shows it is false.

Step-by-Step Solution

Answer

Yes, if \(\sum|a_{n}|\) converges, then \(\sum a_{n}^2\) indeed converges. However, the converse is not true. An example that disproves the converse is the sequence \(a_{n}= (-1)^{n}/\sqrt{n}\).

Step 1: Prove first part using direct proof method

Assume \(\sum|a_{n}|\) is convergent, which means for any \(n\), \(|a_{n}|\) is absolutely convergent. We consider \(|a_{n}|^2\). Since \(|a_{n}|\) is a real number, for all \(n\), |\(a_{n}|\)≤2( \(a_{n}^2\) +1), which implies that \(|a_{n}|^2\)≤2(\(a_{n}^2\) +1). Now, apply the Comparison test. The series \(\sum 2(a_{n}^2\) +1) is convergent because it is the sum of convergent series, so \(\sum a_{n}^2\) is also convergent.

Step 2: Evaluate the converse

We need to find out if the converse of the statement is true or not. The converse being, 'if the series of the squares of the sequence is convergent then the series of the absolute values of the sequence is also convergent'.

Step 3: Find an example that violates the converse

Consider the sequence \(a_{n}= (-1)^{n}/\sqrt{n}\), the series \(\sum a_{n}^2\) is convergent, which can be proved by p-series test where p=2. However, the series of the absolute values \(\sum |a_{n}|\) is equal to \(\sum 1/\sqrt{n}\), which is a divergent harmonic series. This indicates that the converse of the proposition is not true.