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anonymous
 5 years ago
Why is the series (1)^n(n^(1/2)/(n+7)) conditionally convergent? Please help, I have a midterm on this in 4 hours!
anonymous
 5 years ago
Why is the series (1)^n(n^(1/2)/(n+7)) conditionally convergent? Please help, I have a midterm on this in 4 hours!

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0A series converges but diverges absolutely is said to be conditionally convergent.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I realize that. But the problem is when I tested for absolute convergence by taking the absolute value of An >  (n^(1/2))/(n+7) , I found that it An converged. Or maybe I did something incorrectly.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am brushing up on my infinite series, however, if you take the lim as n goes to infinity of n^(1/2)/n it goes to 0 which confirms what you found that it converges

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0^^^ this is not true, just because the limit of terms approaches 0 doesn't mean the series converges.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0In fact n^(1/2)/(2*n) = 1/2 * n^(1/2) < n^(1/2)/n+7) = An By the pseries n^(1/2) diverges and therefore An also diverges

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Great. the n^(1/2) is a clue to manipulate for p series.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry, my computer went crazy

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0YESSS thaanks! Saved me. Im a little more prepared for this test now.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, with these kinds of problems you want to look at the ratio of the ntosomepower terms, in this case we had n^(1/2)/n^1 so n^(1/2), we then realize it must diverge. Because we intend to show it diverges we choose a simpler term that is less than An and show that it diverges, thereby implying An also diverges. In my previous response i used n^(1/2)/(2*n) < n^(1/2)/n+7) which is true after some n and showing that n^(1/2)/(2*n) diverges is far simpler. Obviously if you found that a sum converged then you would find some simple terms Bn > An and then show that sum Bn converges.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes but convince me that 1/2n<1/(n+7)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, I just picked it up: it is true after some n.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can it be that the problem can be written into ((1)^n)*((n^(1/2)/(n+7))) it would totally be easier

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how is that any easier? That was just a couple extra brackets

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks for the review adfriedm. If you don't continue to do these things every day, you forget it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0first of all, the n^(1/2) means that it has to be 0 or positive number. (1)^n means its positive or negitive. if you determine (n^(1/2)/(n+7)) you can see its value decreases and comes to nearly 0. so the series will become a convergent. ps. http://www.wolframalpha.com/input/?i=find+sum+%281%29^n%28n^%281%2F2%29%2F%28n%2B7%29%29+ (you can check the answer here)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We were talking about it being absolutely convergent

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well if you try calculating each number... a(0)=0 a(1)=1/8 a(2)=2^(1/2)/9 a(3)=3^(1/2)/10 a(4)=2/11 ... a(infinity) nearly 0 if you sum them up it will come close to a number. and from the previous link it is nearly 0.0516833.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Just because a(inf) = 0 doesnt mean that the sum of a(n) has a finite value. Have a look at the harmonic series.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh forgot that.. sorry

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait a moment. if a(inf) was nearly 0 then if we sum it up it will be nearly the same number, example s(inf)will nearly be equal to s(inf1) and s(inf+1)will nearly be equal to s(inf) meaning the more we add to it the less it changes proving that this sequence is a convergent
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