Studying the convergence of a series with an arctangent of a partial sum

  • Thread starter Amaelle
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  • #1
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Homework Statement:
look at the image below
Relevant Equations:
asymptotic behaviour, limit of partial sum
Greeting
I'm trying to study the convergence of this serie

1612345260662.png


I started studying the absolute convergence
because an≈n^(2/3) we know that Sn will be divergente S=∝ so arcatn (Sn)≤π/2 and the denominator would be a positive number less than π/2, and because an≈n^(2/3) and we know 1/n^(2/3) > 1/n so we conclude that this serie divergent (not absolutely convergent till this point)

but here is the solution of the book
1612345750480.png

so according to them the serie is absolutely convergent

I would be grateful if someone could explain me where I get wrong in my reasoning

Many thanks in advance!

Best
 

Answers and Replies

  • #2
anuttarasammyak
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You don’t seem to investigate smallness of numerator as the answer does.
 
  • #3
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thanks a lot for your prompt reply
I consider it as a postive number let's call it A 0<A<pi(2) A/n^2/3 > A/n and we know A/n is divergent so should be A/n^2/3 ?
where is the flaw in my reasoning?
 
  • #4
Office_Shredder
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But what is A is really, really, really small?

If A=1/n for example, then A/n does converge.

You assumed there was some constant ##A>0## that was a lower bound for the numerator, but no such constant exists.
 
  • #5
anuttarasammyak
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The solution estimates A as 1/Sn < 1/an
 
  • #6
207
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But what is A is really, really, really small?

If A=1/n for example, then A/n does converge.

You assumed there was some constant ##A>0## that was a lower bound for the numerator, but no such constant exists.
thanks alot
I want to ask about the difference between this case and between when we have for a example a dominator with sin(n) and we just say that |sin(n)|<1
what is the difference between this case and the exercice below?
many thanks in advance!
 
  • #7
Office_Shredder
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Sure. ##|\sin(n)| < 1## is a true statement.

You want to do something like

##|\pi/2 - \arctan(n) | > 0.01##, where 0.01 is a number I just came up with (any number would work). But that's not a true statement, so you can't use it.
 
  • #8
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Sure. ##|\sin(n)| < 1## is a true statement.

You want to do something like

##|\pi/2 - \arctan(n) | > 0.01##, where 0.01 is a number I just came up with (any number would work). But that's not a true statement, so you can't use it.
thanks a million!
 

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