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Extrazlove February 13th, 2019 02:07 PM

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.

idontknow February 13th, 2019 03:19 PM

Yes but the math codes are not showing well .:D

Extrazlove February 13th, 2019 03:24 PM

see message 6

Une suite convergente ? / Enigmes, casse-têtes, curiosités et autres bizarreries / Forum de mathématiques - Bibm@th.net

Extrazlove February 13th, 2019 03:51 PM

Then the sequence is convergent or divergent.
correctly define it in which it is divisive and it contradicts a theorem.

Extrazlove February 14th, 2019 01:25 AM

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

idontknow February 14th, 2019 05:35 AM

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 $\displaystyle \frac{\pi^{2}}{2k^2 }$ .

Extrazlove February 14th, 2019 01:54 PM

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?

Extrazlove February 14th, 2019 01:57 PM

Quote:

Originally Posted by idontknow (Post 605797)
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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