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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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.