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

Clarence Hines

Clarence Hines

Answered

2022-01-24

For an0, and an convergent, show that annp is also convergent for p>12?

Answer & Explanation

Dominique Green

Dominique Green

Expert

2022-01-25Added 11 answers

1). Lemma an, bn0, an, bn convergent, then
anbn
is convergent, too. This is because
anbn12(an+bn)
Now bn=1n2p, bn is convergent
2). a counterexample
an=1n(logn)2
we get
ann=1nlogn
so ann is divergent.
Allison Compton

Allison Compton

Expert

2022-01-26Added 16 answers

Since annp is a sequence with non-negative terms, then the series
n=1annp
converges if and only if it is bounded.
Cauchy-Schwarz provides that
(n=1Nannp)2(n=1Nan)(n=1N1n2p)(n=1an)(n=11n2p)
and as the right-hand side is bounded for 2p>1, then so is the sequence of the partial sums n=1Nannp.
Therefore, the series n=1annp converges for p>12.
RizerMix

RizerMix

Expert

2022-01-27Added 437 answers

If p>1/2, convergence of nan1/2np follows from Cauchy-Schwarz inequality and the fact that nnr is convergence for r>1 If we take an=1n(logn)3/2, then nan is convergent and an1/2n1/2=1n(logn)3/4 and the series n1n(logn)3/4 is divergent.

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