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February 13th, 2019, 03:07 PM   #1
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series vs theorem

Hello

My series is

$R_3=2$
$\displaystyle R_{n+1}=\frac{R_n}{\cos\left(\frac{\pi}{n}\right)}$

I will demonstrate why it is divergent.

To calculate the limit from n to infinity we must first define the n,
so I'm going to calculate it

Can find from the formula that n = pi / arccos (Rn / (Rn + 1)) so Rn / (Rn + 1) is different from 1 to define the n.
The sequence is increasing, so we have (Rn + 1) / Rn > 1 so the result is divergent according to the d'Alembert criterion.


Can demonstrate that this sequence is convergent with the first comparison theorem.
Convergence can be easily demonstrated, but with tools that go beyond high school.

We have for $ n \geq 3 $: $$ R_n = \frac2 {\displaystyle\prod_{k = 3}^{n-1} \cos \left (\frac {\pi} {k} \right)} $$ and the series of logarithms $\displaystyle \sum_{k \geq3} - \ln \left (\cos \left (\frac {\pi} {k} \right) \right) $ converges since its general term is equivalent to $\displaystyle \frac {\pi ^ 2} {2k^2} $. The infinite product converges to a non-zero limit.

Last edited by skipjack; February 14th, 2019 at 07:42 AM. Reason: to improve markup
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February 13th, 2019, 04:19 PM   #2
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Yes but the math codes are not showing well .
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February 13th, 2019, 04:24 PM   #3
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see message 6

Une suite convergente ? / Enigmes, casse-têtes, curiosités et autres bizarreries / Forum de mathématiques - Bibm@th.net
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February 13th, 2019, 04:51 PM   #4
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Then the sequence is convergent or divergent.
correctly define it in which it is divisive and it contradicts a theorem.
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February 14th, 2019, 02:25 AM   #5
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look at this discussion to understand(us google traduction).
In mathematics or computer science, you must define the variables before using them (n Rn x ....)
In the 2 demonstration he plays with an indeterminate form of n so that demonstration is false.
Une suite convergente ? / Enigmes, casse-têtes, curiosités et autres bizarreries / Forum de mathématiques - Bibm@th.net
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February 14th, 2019, 06:35 AM   #6
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What does it mean and how to get there or prove it ?
$\displaystyle \sum_{k\geq 3} -\ln(\cos (\frac{\pi}{k}))\; $ converges since the general term is equivalent to $\displaystyle \frac{\pi^{2}}{2k^2 }$ .
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February 14th, 2019, 02:54 PM   #7
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Here is a bizarre series I will show that it diverges and normally the comparison theorem shows who it converges is equivalent to $\displaystyle \frac{\pi^{2}}{2k^2 }$.
R3 = 2

Rn + 1 = rn / cos (pi / n)
I will demonstrate who she is divergent

To calculate the limit of n to infinity, we must first define the n for not falling on things that do not exist like 1/0.

So I'm going to calculate it

In Can find from the formula that n = PI / arcos (Rn / Rn + 1) So Rn / Rn + 1 is different from 1 to define well the n and not to fall on an absurdity 1/0 which leads to false calculations .

The sequence is thus increasing Rn + 1 / Rn> 1 so the result is divergent according to the Dalembert criterion so as not to fall on a 1/0 absurdity.


Who is the valid demonstration?
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February 14th, 2019, 02:57 PM   #8
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Quote:
Originally Posted by idontknow View Post
What does it mean and how to get there or prove it ?
$\displaystyle \sum_{k\geq 3} -\ln(\cos (\frac{\pi}{k}))\; $ converges since the general term is equivalent to $\displaystyle \frac{\pi^{2}}{2k^2 }$ .

It is a famous mathematician who has dismantled it here
Une suite convergente ? / Enigmes, casse-têtes, curiosités et autres bizarreries / Forum de mathématiques - Bibm@th.net
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