I was reading this question on this website.
Let {an} be a sequence of non-negative real numbers such that
the series ∑∞n=1an is convergent.
If p is a real number such that the series ∑√annp diverges, then
(A) p must be strictly less than 12
(B) p must be strictly less than or equal to 12
(C) p must be strictly less than or equal to 1 but can be greater
than 12
(D) p must be strictly less than 1 but can be greater than or equal
to 12.
I spent a lot of time thinking about the problem but after lot of mental struggle I gave up. I looked at the answers posted and it was a simple application of Cauchy-Schwarz inequality.
Being a bit more explicit, the Cauchy-Schwarz inequality and the
assumptions imply
∞=∞∑n=1a1/2nnp≤(∞∑n=1an)1/2(∞∑n=11n2p)1/2.
In short my question is what are some red flags to notice which signals that Cauchy-Schwarz inequality can be used to solve the problem.
I know the theorem and proof as well.
P.S. Sorry for my English. Feel free to edit.
Answer
You've got an arbitrary convergence series ∑an, and you're trying to figure out the convergence of ∑√annp. Two things strike me about that series.
Firstly, the terms are related to √an, rather than just an. That will change convergence, in particular, it will make convergence slower (or make convergence not happen at all). If we want to relate this series back to ∑an, it'd be really handy if there was some inequality that naturally involved looking at the sum of squared terms, such as the Cauchy-Schwarz inequality.
Secondly, we're taking two simple series, and multiplying them term-by-term, sort of like an infinite dot product. Working with that is hard. If there was an inequality that could separate the term-by-term product of series, we might have something to work with. The Cauchy-Schwarz inequality might help there.
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