anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Never heard it called the absolute convergence test. Are you talking about the ratio test?
anonymous
  • anonymous
\[\lim_{n \rightarrow \infty} \left| a(n+1)/a(n) \right|\] ?
anonymous
  • anonymous
yes haha

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

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