## cherrilyn 5 years ago PROOF! of the sum a sub n........

1. cherrilyn

Prove that if $\sum_{}^{}a _{n}$ converges absolutely then $\sum_{}^{} a _{n}^{2}$ also converges. Then show by giving a counterexample that $\sum_{}^{} a _{n}^{2}$ need not converge if $\sum_{}^{}a _{n}$ is only conditionally convergent.

2. watchmath

How rigorous do you want the proof to be? Is it using epsilon-delta type of proof?

3. cherrilyn

no..we just learned about harmonic, p-series, alternating harmonic...absolute and conditional convergence....... so I don't think that rigorous

4. watchmath

Well, we can do the following: Since the first series is convergent, then there is an \M\) such that $$0\leq |a_n|<1$$ for all $$n\geq M$$ (think that if there are infinitely many terms that greater than 1, then the series diverges). But then $$a_n^2<a_n$$ for all $$n\geq M$$ Hence by the comparison test the second series is convergent as well.

5. watchmath

Example: $\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}$ is is conditionally convergent but $\sum_{n=1}^\infty \frac{1}{n}$is divergent.