## 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.

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.