Question

PROOF! of the sum a sub n........

Answer 1

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.

Answer 2

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

Answer 3

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

Answer 4

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.

Answer 5

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