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Study23
 one year ago
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 \(\ \frac{n}{n+5} \).
Study23
 one year ago
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 \(\ \frac{n}{n+5} \).

This Question is Closed

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

tkhunny
 one year ago
Best ResponseYou've already chosen the best response.0First, 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.
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