The series $\sum_n^\infty a_n^p$ where $\{a_n\}_{n=1}^\infty$ is a convergent, strictly positive sequence

Question

Suppose that $\{a_n\}_{n=1}^\infty$ is a sequence of strictly positive numbers and that $\sum_n^\infty a_n=A$ is a convergent series. Suppose that $p >1$.What can you say about the series $\sum_n^\infty a_n^p$?

Because $a_k>0 \, \forall k$ and its series converges, we know that $\{a_n\}$ is absolutely convergent. We can rearrange the sequence of terms in an absolutely convergent sequence, so let $\{q_n\}$ be $\{a_n\}$ in reverse.

Now, we know that if $\sum_n^\infty a_n=A$ converges absolutely and $\sum_n^\infty b_n=B$ converges, then $$\sum_{n=0}^\infty \sum_{k=0}^n a_n b_{n-k} = AB.$$

I thought that I might be able to use the absolute convergence of $\sum a_n$ to "recursively" build to $a_n^p$ through repeated multiplication of $\{q_n\}$ and $\{a_n\}$, but I'm not sure how to implement this idea in the face of the double summation.

Is this approach the best one? If so, where do I go from here? If not, how should I have approached the problem?

Answer 1

Since $\sum a_n$ converges you have $a_n \to 0$ so that $M = \sup a_n$ is finite. Then $$\sum a_n^p \le \sum M^{p-1} a_n = M^{p-1} \sum a_n < \infty.$$

Moreover, since $M \le \sum a_n$ you also have $$ \sum a_n^p \le \left( \sum a_n \right)^p.$$

Answer 2

Note that $a_n\stackrel{n \to \infty}{\longrightarrow} 0$ as $\sum a_n <\infty$.

According to limit comparison test you have

$$\frac{a_n^p}{a_n}=a_n^{p-1}\stackrel{p > 1, n \to \infty}{\longrightarrow} 0 \stackrel{\sum a_n <\infty}{\Longrightarrow}\sum a_n^p < \infty$$