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Question: If $\sum a_n$ is convergent, with $a_n \geq 0$ $\forall n \in \mathbb{N}$, show that if $p > 1$, $\sum a_n^p$ converges.

Here's my proof. Can someone please verify it?

$\displaystyle{\left( \sum\limits_{k=1}^n a_k \right)^p \geq \sum\limits_{k=1}^n a_k^p} \geq 0$

Note that $\displaystyle{\left( \sum\limits_{k=1}^n a_k \right)^p}$ is bounded, since the sequence of partial sums $\sum\limits_{k=1}^n a_k$ converges. Also, the sequence $s_n = \sum\limits_{k=1}^n a_k^p$ is increasing. Therefore, by the monotone convergence theorem, $s_n$ converges.

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  • $\begingroup$ Your proof seems fine to me. $\endgroup$ – Samrat Mukhopadhyay May 14 '14 at 9:21
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As $\sum a_n$ converges, there is an $N \in \mathbb{N}$ such that $$\forall n \geq N \implies |a_n| < 1.$$

As $a_n \geq 0$, we have $$a_n^2 < a_n<1.$$

Thus by induction we get, $$a_n^p < a_n.$$

The result now follows from the Comparison Test.

Remark: Notice as $a_n \geq 0$, absolute convergence also sneaked in.

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