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Study23
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Okay. I'm having trouble with series. Here's the problem (I'm looking to see whether the series is convergent or divergent, if convergent I have to find the sum it converges to): summation from n=1 to infinity of \(\ \frac{n}{n+5} \).
 one year ago
 one year ago
Study23 Group Title
Okay. I'm having trouble with series. Here's the problem (I'm looking to see whether the series is convergent or divergent, if convergent I have to find the sum it converges to): summation from n=1 to infinity of \(\ \frac{n}{n+5} \).
 one year ago
 one year ago

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Study23 Group TitleBest ResponseYou've already chosen the best response.0
I'm thinking partial fractions or factoring n out but I'm not sure...
 one year ago

macknojia Group TitleBest ResponseYou've already chosen the best response.1
can you type this out i can't read it. Sorry :(
 one year ago

macknojia Group TitleBest ResponseYou've already chosen the best response.1
nvm, i got it. You can view this as a geometric series.\[n/(n(5))\]
 one year ago

macknojia Group TitleBest ResponseYou've already chosen the best response.1
so your first term is n and your ratio is 5. now using the formula you can find the sum it converges to.
 one year ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
So you factor the n out?
 one year ago

macknojia Group TitleBest ResponseYou've already chosen the best response.1
not really. are you familiar with the formula to find the sum of a convergent series ?
 one year ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
I meant to find what a and r is before I find the sum.
 one year ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
Because r is less than 1 (5) does that mean this is divergent?
 one year ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.0
What is the limit of the TERMS as n increases without bound? If it's not zero, it doesn't converge.
 one year ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
But r has to be 1<r<1 right?
 one year ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
r is 5 so...
 one year ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.0
Yes, that is good for 'r', but you don' even need that. Just look at the limit of the terms. Test #1, in ALL cases.
 one year ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.0
BTW, r is NOT 5. It's 1.
 one year ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
. ? And wouldn't the limit as n approaches infinity be 0 because denominator is larger?
 one year ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.0
No. The limit of the terms is 1. As n increases, the 5 becomes insignificant.
 one year ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
Okay. And how do I find r to be one then?
 one year ago

macknojia Group TitleBest ResponseYou've already chosen the best response.1
the coefficient of numerator and the denominator is 1 therefore, the limit is one.
 one year ago

macknojia Group TitleBest ResponseYou've already chosen the best response.1
try graphing it.
 one year ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.0
First, I don't actually care. Since the limit of the terms is 1, it cannot converge. There is no reason to do the ration test. Second, just do it. \(\dfrac{\dfrac{n+1}{n+6}}{\dfrac{n}{n+5}} = \dfrac{n^{2}+6n+5}{n^{2}+6n}\). The limit of this, as n increases without bound, is 1. As n increases, 5 becomes insignificant and it is clearly 1.
 one year ago
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