**Problem Statement:** Suppose that converges absolutely. Prove that converges for every .

**Proof:** We are given that converges absolutely, this implies that converges. By definition this means that the sequence of partial sums, , converges. But every convergent sequence is also Cauchy.

Let and let . Since is Cauchy there is an such that for every . We wish to show that converges by showing that it’s Cauchy, that is, that for .

Consider multiplied times. Each of the is < for . This gives us the following inequality when : since . This implies that for every , thus is Cauchy. Thus converges and converges.

**Reflection: **The meat of this proof is in the Cauchy criterion for series, which allowed us to show the partials of the series converged, giving us that the series itself converges.

Or, if the sum converges, then the sequence {a_k} -> 0.

So, fix N s.t. k>N implies |a_k| < 1.

Hence |a_k|^p < |a_k|.

So by the comparison test, the sum of |a_k|^p converges.

Q.E.D.

Cauchy ftw! I do agree with allen though; as much as I love Cauchy, I would use the comparison test.

Me too. Comparison all the way.