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A series converges but diverges absolutely is said to be conditionally convergent.
I 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.
I 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
^^^ this is not true, just because the limit of terms approaches 0 doesn't mean the series converges.
In fact n^(1/2)/(2*n) = 1/2 * n^(-1/2) < n^(1/2)/n+7) = |An| By the p-series n^(-1/2) diverges and therefore |An| also diverges
Great. the n^(1/2) is a clue to manipulate for p series.
Sorry, my computer went crazy
YESSS thaanks! Saved me. Im a little more prepared for this test now.
Yeah, with these kinds of problems you want to look at the ratio of the n-to-some-power 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.
Yes but convince me that 1/2n<1/(n+7)
Oh, I just picked it up: it is true after some n.
can it be that the problem can be written into ((-1)^n)*((n^(1/2)/(n+7))) it would totally be easier
how is that any easier? That was just a couple extra brackets
Thanks for the review adfriedm. If you don't continue to do these things every day, you forget it.
first 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+%28-1%29^n%28n^%281%2F2%29%2F%28n%2B7%29%29+ (you can check the answer here)
We were talking about it being absolutely convergent
well 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.
Just because a(inf) = 0 doesnt mean that the sum of a(n) has a finite value. Have a look at the harmonic series.
oh forgot that.. sorry
wait 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(inf-1) 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