- Provide examples to show that the series \(\sum_{n=1}^{\infty} a_n^2\) may or may not converge when the series \(\sum_{n=1}^{\infty} a_n\) converges conditionally.
- Prove that if the series \(\sum_{n=1}^{\infty} a_n\) converges absolutely, then the series \(\sum_{n=1}^{\infty} a_n^2\) must converge.
Solution
1. Conditional convergence of \(\sum a_n\) can occur while \(\sum a_n^2\) converges.
Take \[ a_n=\frac{(-1)^{n+1}}{n}. \] Then \(\sum a_n\) converges by the alternating-series test, but not absolutely because \(\sum 1/n\) diverges. Thus \(\sum a_n\) converges conditionally. On the other hand, \[ \sum_{n=1}^\infty a_n^2=\sum_{n=1}^\infty \frac1{n^2} \] converges.
2. Conditional convergence of \(\sum a_n\) can also occur while \(\sum a_n^2\) diverges.
Take \[ a_n=\frac{(-1)^{n+1}}{\sqrt n}. \] The series \(\sum a_n\) converges by the alternating-series test because \(1/\sqrt n\) decreases to \(0\), but it is not absolutely convergent because \(\sum 1/\sqrt n\) diverges. Hence \(\sum a_n\) converges conditionally. However, \[ \sum_{n=1}^\infty a_n^2=\sum_{n=1}^\infty \frac1n \] diverges.
3. If \(\sum a_n\) converges absolutely, then \(\sum a_n^2\) converges.
Absolute convergence means \[ \sum_{n=1}^\infty |a_n|<\infty. \] Therefore \(a_n\to0\), so there exists \(N\) such that \(|a_n|\le1\) for all \(n\ge N\). For such \(n\), \[ 0\le a_n^2=|a_n|^2\le |a_n|. \] Thus \(\sum_{n=N}^\infty a_n^2\) converges by comparison with \(\sum_{n=N}^\infty |a_n|\). Adding the finitely many initial terms gives convergence of \(\sum_{n=1}^\infty a_n^2\).