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

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The discussion focuses on solving the infinite series involving factorials and pi, specifically the series 1/2! - (π²/4²4!) + (π⁴/4⁴6!). Participants suggest that the series resembles the Maclaurin expansion of cos(x) and explore substituting x = π/4. They identify that the factorials in the series increase by 2, leading to the conclusion that the series can be expressed as F(π/4), where F(x) is defined in terms of a summation. By manipulating the series and comparing terms, they derive a relationship that simplifies to s = (4/π)²(1 - cos(π/4)). The conversation highlights the importance of matching factorials and powers to solve such series effectively.
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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}})
 
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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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