Convergent Alternating Series. Calculus. Help please.
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Convergent Alternating Series. Calculus. Help please.

[From: ] [author: ] [Date: 12-11-06] [Hit: ]
So, we need (0.6)^(2n-1) / (2n-1)!==> n = 5 (or higher), by inspection.I hope this helps!......
I'm confused. Any type of help will be great.

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The nth term (in absolute value) is (0.6)^(2n-1) / (2n-1)!.

Since we have an alternating series, the error after n terms is bounded above by the absolute value of the (n+1)th term:

So, we need (0.6)^(2n-1) / (2n-1)! < 0.0000001 = 10^(-7),
==> n = 5 (or higher), by inspection.

I hope this helps!
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