Drsoccerball
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- 2015
Yes, this is correct. What did you compare the first series to though?Thanks guys just a question from the exam:
I said the first one diverges by comparison test and by Leibniz test the second one conditionally converges. Am I right?
I had like a few seconds so i just checked it with the harmonic series without checking the inequality sign... R.I.PYes, this is correct. What did you compare the first series to though?
Here's the Wikipedia page for the divergence of the sum of reciprocals of primes (which has a few proofs of it there), if you (or anyone else) want(s) to see it: https://en.wikipedia.org/wiki/Divergence_of_the_sum_of_the_reciprocals_of_the_primes .I had like a few seconds so i just checked it with the harmonic series without checking the inequality sign... R.I.P
Wrote the question wrong ahah :'(Here's the Wikipedia page for the divergence of the sum of reciprocals of primes (which has a few proofs of it there), if you (or anyone else) want(s) to see it: https://en.wikipedia.org/wiki/Divergence_of_the_sum_of_the_reciprocals_of_the_primes .
Did they get you to derive some bounds on p_n or something? It's not exactly a trivial thing to prove, that divergence.
Hahaha was it π(n) that the Q. was actually referring to?Using p_n for pi(n) is horrific notation, but it remains true that the first one diverges and the second one converges, and it remains true that proving divergence is nontrivial (quite similar though).
What I wrote was what the queation gaveHahaha was it π(n) that the Q. was actually referring to?
Edit: oh, just noticed Drsoccerball's post above. (Wasn't able to see the LaTeX before, I think it's playing up for me at the moment.)
Did the Q. provide any fact about the asymptotic nature of π(n)?
https://en.wikipedia.org/wiki/Prime-counting_functionWhat I wrote was what the queation gave
From the wording, they probably just wanted you do decide whether:Did the Q. provide any fact about the asymptotic nature of π(n)?