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)!
∑_(n=0)^∞▒((-1)^2 (n!)^2)/(2n)!
-
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.
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.
-
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
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