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anonymous
 5 years ago
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..)
anonymous
 5 years ago
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Never heard it called the absolute convergence test. Are you talking about the ratio test?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{n \rightarrow \infty} \left a(n+1)/a(n) \right\] ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ah 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
 5 years ago
Best ResponseYou've already chosen the best response.0But you can bypass that step by multiplying by the reciprocal once you get the hang of it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yup got that, i'm having trouble simplifying that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Your k! when your replacing "k" with "k+1" will end up being "(k+1)!"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is that what you meant or as you progress farther into the problem when you have k! over (k+1)!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0further, taking the limit of the already established reciprocal

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok. And what exact thing are you having trouble multiplying?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right now i have 1^(k+1)+1/(k+1)! * k!/1^(k+1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so whats next? how do i find that limit/simplifiy that fraction

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok. 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
 5 years ago
Best ResponseYou've already chosen the best response.0Let me look in my notes real quick

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cause i'll also need that rule to test the original function for convergence

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok here is what happens. Lets say you have n! over (n+1)!. Your result will be (n+1) in the denominator (no factorial)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The factorials cancel each other in a way.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0..and you'll keep the constants right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well 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
 5 years ago
Best ResponseYou've already chosen the best response.0So the first term in what you have right now has a (k+1) in the denominator instead of (k+1)!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so i'll end up taking the limit of 1/k+1 ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks for your help! :)
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