(January 2003 #1) Let \(\{a_k\}\) be a sequence of real numbers such that the series \(\sum_{k=1}^{\infty}a_k\) converges and \(\sum_{k=1}^{\infty}a_k^2\) diverges. Prove that \(\sum_{k=1}^{\infty}a_k\) does not converge absolutely. (See also June 2010 #3a where you are instead told that \(\sum_{k=1}^{\infty}a_ka_{k+1}\) diverges and asked to show the same result. Compare this to June 2009 #3a and January 2005 #1b.)
Solution
1. Suppose, toward a contradiction, that \(\sum |a_k|\) converges.
2. Then \(|a_k| \le 1\) eventually.Proof: convergence of \(\sum |a_k|\) forces \(|a_k| \to 0\), so for all large \(k\), \(|a_k| \le 1\), hence \(a_k^2 \le |a_k|\).
3. \(\sum a_k^2\) converges, contradiction.Proof: for all large \(k\), \(0 \le a_k^2 \le |a_k|\); by comparison with the convergent series \(\sum |a_k|\), the series \(\sum a_k^2\) converges — contradicting the hypothesis that it diverges.
4. Q.E.D. Proof: hence \(\sum a_k\) does not converge absolutely.