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

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?

• What does "$\{a_n\}$ in reverse" mean? – 5xum Oct 18 '18 at 12:17
• Hint: You can notice that $a_n \to 0$. Let $N$ s.t $a_n<1$ $\forall n >N$. Then as $p>1$, for all $n>N$ you have $a_n^p \leq a_n$. – Delta-u Oct 18 '18 at 12:17

## 2 Answers

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 $$\sum a_n^p \le \left( \sum a_n \right)^p.$$

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

According to limit comparison test you have

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