What is the solution to the infinite series involving factorials and pi?

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Homework Help Overview

The discussion revolves around an infinite series involving factorials and powers of pi, specifically the series given by the original poster. Participants are exploring connections to known series expansions, particularly the Maclaurin series for cosine.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss attempts to relate the series to the Maclaurin expansion of cosine by substituting specific values for x. There is a focus on matching terms based on factorials and powers of pi, with some questioning how to manipulate the factorials to align with the original series.

Discussion Status

Several participants are actively engaging with the problem, offering insights and suggestions for approaching the series. There is a recognition of the factorial structure and its implications, with some participants proposing to multiply the series by specific terms to explore further relationships. No consensus has been reached, but the discussion is productive.

Contextual Notes

Participants are navigating the complexities of factorial manipulation and series convergence, with some expressing uncertainty about their previous assumptions and calculations. The original series setup and the nature of the problem are under examination.

lionely
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Homework Statement



Sum to infinity

\frac{1}{2!} - \frac{ \pi ^2}{4^2.4!} + \frac{\pi^4}{4^4.6!} ...

Homework Equations





The Attempt at a Solution



I thought the series was similar to the Maclaurin expansion of cos x

so I tried putting in x= ∏/4

But I end up with the series {1} - \frac{ \pi ^2}{4^2.2!} + \frac{\pi^4}{4^4.4!} ...

I don't know how to change the factorials to make it match the ones in the series given.
 
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Try matching up the terms of the two series based on the factorials, not on the powers of pi. What do you notice?
 
Well the factorials increase by 2. so it's like (2n)!?
 
You may want to look a bit closer at how you factored 1/2! out.

4! = 2! *3*4 <> 2*2!
 
I know 4! is not equal to 2 x 2!. I didn't factor out anything all I did was put in x = pi/4 in the Maclaurin expansion of CosX
 
lionely said:

Homework Statement



Sum to infinity

\frac{1}{2!} - \frac{ \pi ^2}{4^2.4!} + \frac{\pi^4}{4^4.6!} ...

You can write this as F(\pi/4), where
<br /> F(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+2)!}x^{2n}<br />
so that
<br /> x^2F(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+2)!}x^{2n+2}<br /> = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2(n+1))!}x^{2(n+1)}<br />
 
Multiply the whole series by (π/4)2 and see what you get!
 
lionely said:
Well the factorials increase by 2. so it's like (2n)!?
Yes. Compare the term with (2n)! in the first series with the term with (2n)! in the second series, same n.
 
s= \frac{1}{2!} - \frac{ \pi ^2}{4^2.4!} + \frac{\pi^4}{4^4.6!} ...

(\frac{π}{4})^2s= \frac{(\frac{π}{4})^2}{2!}-\frac{(\frac{π}{4})^4}{4!}+\frac{(\frac{π}{4})^6}{6!} ... =1-\cos{\frac{π}{4}}

s=(\frac{4}{π})^2(1-\cos{\frac{π}{4}})=(\frac{4}{π})^2(1-\frac{1}{\sqrt{2}})
 
  • #10
Wow Chestermiller, I feel kind of stupid now. Thank you... I was thinking that it looked like cosx but the factorials weren't adding up, I should of tried to get the powers up so it could match. Thank you again guys.
 

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