September 2001 #2

Problem Statement: Suppose that \sum_{k=1}^\infty a_k converges absolutely. Prove that \sum_{k=1}^\infty|a_k|^p converges for every p\geq 1.

Proof: We are given that \sum_{k=1}^\infty a_k converges absolutely, this implies that \sum_{k=1}^\infty |a_k| converges. By definition this means that the sequence of partial sums, \{S_n= \sum_{k=1}^n |a_k|\}, converges. But every convergent sequence is also Cauchy.

Let p\geq 1 and let \varepsilon >0. Since \{S_n\} is Cauchy there is an N\in \mathbb{N} such that \left|\sum_{k=n}^m|a_k|\right|<\left(\dfrac{\varepsilon}{p}\right)^{1/p} for every m>n\geq N.  We wish to show that \{T_n=\sum_{k=1}^n|a_k|^p\} converges by showing that it’s Cauchy, that is, that \left|\sum_{k=n}^m|a_k|^p\right|<\varepsilon for m>n\geq N.

Consider \left|\sum_{k=n}^m|a_k|^p\right|\leq \left|\sum_{k=n}^m|a_k|\right|\dots\left|\sum_{k=n}^m|a_k|\right| multiplied p times. Each of the \left|\sum_{k=n}^m|a_k|\right| is < \left(\dfrac{\varepsilon}{p}\right)^{1/p} for m>n\geq N. This gives us the following inequality when m>n\geq N|\sum_{k=n}^m|a_k|^p|<\left(\left(\dfrac{\varepsilon}{p}\right)^{1/p}\right)^p=\dfrac{\varepsilon}{p}\leq \varepsilon since p\geq 1. This implies that |\sum_{k=n}^m|a_k|^p|<\varepsilon for every m>n\geq N, thus \{T_n\} is Cauchy. Thus \{T_n\} converges and \sum_{k=1}^\infty|a_k|^p converges.

\Box

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.   

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This entry was posted in Analysis, Cauchy, Sequence, Series and tagged , , . Bookmark the permalink.

3 Responses to September 2001 #2

  1. Allan Boone says:

    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.

  2. Richard says:

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

  3. DR says:

    Me too. Comparison all the way.

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