10 pts. calculus help please
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10 pts. calculus help please

[From: ] [author: ] [Date: 12-04-24] [Hit: ]
/(n!And since 1/4 -Somehow I dont think youve put the right problem in the question.Summation from n=0 - infinity of f(x) = (-1^2*(n!)^2)/(2n)!Which does not converge to 1/4. basically,......
Me and my girlfriend have been starring at this problem. Our teacher told us it converges absolutely to 1/4. how do you do it?

∑_(n=0)^∞▒((-1)^2 (n!)^2)/(2n)!

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You take the absolute value of the term, and use the ratio test

limit n->∞ (n+1)!^2/(2n+2)! * (2n)!/(n!)^2

limit n->∞ (n+1)^2/[(2n+2)(2n+1)]

limit n->∞ (n^2+2n+1)/(4n^2+6n+2)

1/4

And since 1/4 < 1, the series is absolutely convergent.

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Somehow I don't think you've put the right problem in the question. I read it as follows:

Summation from n=0 - infinity of f(x) = (-1^2*(n!)^2)/(2n)!

Which does not converge to 1/4. basically, what I wrote reduces as follows: -1^2 = 1, so (n!)^2/(2n)!

(n!)^2/(2*(n!))

Cancel an n!

n!/2 which diverges by itself, so the sum certainly does.

So, if you clarify the problem I'll look at it again, or possible the way I reduced it gives you a clue of what to do.

Thinkingblade
1
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