My Math Forum series vs theorem

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 February 13th, 2019, 02:07 PM #1 Banned Camp   Joined: Feb 2019 From: Casablanca Posts: 23 Thanks: 2 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 06:42 AM. Reason: to improve markup
 February 13th, 2019, 03:19 PM #2 Senior Member   Joined: Dec 2015 From: somewhere Posts: 605 Thanks: 88 Yes but the math codes are not showing well .
 February 13th, 2019, 03:24 PM #3 Banned Camp   Joined: Feb 2019 From: Casablanca Posts: 23 Thanks: 2 Thanks from idontknow
 February 13th, 2019, 03:51 PM #4 Banned Camp   Joined: Feb 2019 From: Casablanca Posts: 23 Thanks: 2 Then the sequence is convergent or divergent. correctly define it in which it is divisive and it contradicts a theorem.
 February 14th, 2019, 01:25 AM #5 Banned Camp   Joined: Feb 2019 From: Casablanca Posts: 23 Thanks: 2 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 Thanks from idontknow
 February 14th, 2019, 05:35 AM #6 Senior Member   Joined: Dec 2015 From: somewhere Posts: 605 Thanks: 88 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 $\displaystyle \frac{\pi^{2}}{2k^2 }$ .
 February 14th, 2019, 01:54 PM #7 Banned Camp   Joined: Feb 2019 From: Casablanca Posts: 23 Thanks: 2 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?
February 14th, 2019, 01:57 PM   #8
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Quote:
 Originally Posted by idontknow 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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