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let $a_n$ be a sequence of real numbers such that the series $\sum |a_n|^2$ is convergent.Find range of $p$ such that the series $\sum |a_n|^p$ is convergent.

My try:

To show the series it is necessary to show that sequence of partial sums of $\sum |a_n|^p$ is bounded.

Consider $S_n=|a_1|^p+|a_2|^p+...+|a_n|^p$. How to proceed next? Any inequality which can be used?

amWhy
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    Comparison says that since $;|a_n|<1;$ for almost all $;n;$ (why?), then $;|a_n|^p\le |a|^2;$ if $;p\ge2;$ , so then --- – Timbuc Jan 28 '15 at 11:47
  • With the information given, you cannot find the set of all $p$, such that the series is convergent. That is, you won't be able to tell what happens for $p<2$. – Mankind Jan 28 '15 at 11:51

3 Answers3

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If you denote $M = \max\{|a_n|\}$ you have $$\sum_{n=1}^\infty |a_n|^p = \sum_{n=1}^\infty |a_n|^{p-2} |a_n|^2 \le M^{p-2} \sum_{n=1}^\infty |a_n|^2 < \infty$$whenever $p \ge 2$.

Umberto P.
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If $p\geq2$ then $\sum|a_n|^p$ converges since $$\lim\limits_{n\to\infty}\frac{|a_n|^p}{|a_n|^2}=\lim\limits_{n\to\infty}|a_n|^{p-2}=0$$ where p>2. On the other hand if $0<p<2$ then there exists a sequcence such that $\sum|a_n|^2$ is convergent but $\sum|a_n|^p$ is divergent. Formally, for all $p\in(0,2)$ there exists $0<\varepsilon<\frac{2}{p}-1$. then if we choose the sequence $a_n=\frac{1}{n^{\frac{1+\varepsilon}{2}}}$ then $\sum|a_n|^2$ converges but $\sum|a_n|^p$ does not converge.

guest
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Suppose that $\ell_q$ is the space of real sequences $x=\{x_n\}$ such that $\sum|x_n|^q<\infty$ with the norm $$ \|x\|_q=\biggl(\sum|x_n|^q\biggr)^{1/q} $$ for $q\ge1$. We have that $\|x\|_p\le\|x\|_q$ for $1\le q\le p<\infty$ (see here). Hence, the series $\sum|a_n|^p$ converges for all $p>2$.

To see that the series might not necessarily converge for $p<2$, let's take, for example, $a_n=n^{-1/p}$ with $p<2$.

Cm7F7Bb
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