For an≥0, and ∑an convergent, show that ∑annp is also convergent for p&gt;12?

Question

For an≥0, and ∑an convergent, show that ∑annp is also convergent for p&amp;gt;12?

Dominique Green · Accepted Answer

1). Lemma an, bn≥0, ∑an, ∑bn convergent, then  ∑anbn  is convergent, too. This is because  anbn≤12(an+bn)  Now bn=1n2p, ∑bn is convergent  2). a counterexample  an=1n(log⁡n)2  we get  ann=1nlog⁡n  so ∑ann is divergent.

Allison Compton · Accepted Answer

Since annp is a sequence with non-negative terms, then the series  ∑n=1∞annp  converges if and only if it is bounded.  Cauchy-Schwarz provides that  (∑n=1Nannp)2≤(∑n=1Nan)(∑n=1N1n2p)≤(∑n=1∞an)(∑n=1∞1n2p)  and as the right-hand side is bounded for 2p&amp;gt;1, then so is the sequence of the partial sums ∑n=1Nannp.  Therefore, the series ∑n=1∞annp converges for p&amp;gt;12.

RizerMix · Accepted Answer

If p&amp;gt;1/2, convergence of ∑nan1/2n−p follows from Cauchy-Schwarz inequality and the fact that ∑nn−r is convergence for r&amp;gt;1 If we take an=1n(log⁡n)3/2, then ∑nan is convergent and an1/2n−1/2=1n(log⁡n)3/4 and the series ∑n1n(log⁡n)3/4 is divergent.

For a_n\geq0, and \sum a_n convergent, show that \sum\frac{\sqrt{a_n}}{n^p} is

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