determine if the summation to infinity of the alternating sum is absolutely convergent, conditionally convergent or divergent: (-1)^(k+1)/k! (having problems simplifying the absolute convergence test, for some reason i was never taught how to simplify the factorial when it was in a fraction..)

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determine if the summation to infinity of the alternating sum is absolutely convergent, conditionally convergent or divergent: (-1)^(k+1)/k! (having problems simplifying the absolute convergence test, for some reason i was never taught how to simplify the factorial when it was in a fraction..)

Mathematics
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Never heard it called the absolute convergence test. Are you talking about the ratio test?
\[\lim_{n \rightarrow \infty} \left| a(n+1)/a(n) \right|\] ?
yes haha

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Ah ok. Well what your going to do is you write your function out, replacing "n" with "n+1". Then your going to divide it by your function written with just "n" for your variable.
But you can bypass that step by multiplying by the reciprocal once you get the hang of it.
yup got that, i'm having trouble simplifying that
Your k! when your replacing "k" with "k+1" will end up being "(k+1)!"
Is that what you meant or as you progress farther into the problem when you have k! over (k+1)!
further, taking the limit of the already established reciprocal
multiplaction step
ok. And what exact thing are you having trouble multiplying?
finding the limit
right now i have 1^(k+1)+1/(k+1)! * k!/1^(k+1)
so whats next? how do i find that limit/simplifiy that fraction
Ok. there is a property that comes into play when you have n!/(n+1)!. I'm trying to think of how to show it. It reduces it down.
Let me look in my notes real quick
ok :)
cause i'll also need that rule to test the original function for convergence
Ok here is what happens. Lets say you have n! over (n+1)!. Your result will be (n+1) in the denominator (no factorial)
The factorials cancel each other in a way.
..and you'll keep the constants right?
Well the only thing affected in what I just told you is your k! and (k+1)! The rest of the problem reduces just like typical multiplication & cancelling.
So the first term in what you have right now has a (k+1) in the denominator instead of (k+1)!
so i'll end up taking the limit of 1/k+1 ?
I believe so
wonderful :)
thanks for your help! :)
Your very welcome

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