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anonymous
 5 years ago
sigma (k=1) to infinity (1)^(k+1) sqrt(k)/k+1 . I tried to solve it using the ratio test, but got stuck in the end, and I can't use root test since the whole thing is not to the power of 1/k. Which way can I use?
anonymous
 5 years ago
sigma (k=1) to infinity (1)^(k+1) sqrt(k)/k+1 . I tried to solve it using the ratio test, but got stuck in the end, and I can't use root test since the whole thing is not to the power of 1/k. Which way can I use?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{(k+1)\sqrt(k+1)}{k^{3/2} + 2 \sqrt(k)}\] that's where I got stuck after using the ratio test

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{k=1}^{\infty}(1)^{k+1} \frac{\sqrt(k)}{k+1}\] that's the question

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh wow, alas, visitors lol ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It doesn't say what you are doing? test for convergence?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol, yes ^_^ but I'm not sure which way to choose, I tried ratio test

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0umm it looks like you could maybe use divergence test

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0kth term divergence test? ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so from where I stopped , I can use the kth term divergence test? right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok with divergence test you get = 0 so that is inconclusive

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no, so that means it's convergent!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but, the book says it's conditionally convergent . That's where I'm confused :(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh I see now! The series is increasing, but it's limit tends to zero, so that's why it's conditonally convergent right? ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay so when you get 0 from the divergence test, that tells you to try another test basically. And if you can get it to converge with another test (that is not absolute convergence) then its conditionally convergent

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0um, that's a good guess. But I don't think so

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so with the ratio test I got 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so again, that means nothing

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0um so what other tests are there

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0AST, pseries, root test, geometric series

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0isn't my logic right lol? the series must be decreasing to be absolutely convergent, but it's increasing instead

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm not sure enough to say yes or no to that. See what happens when you do more conditionally convergent problems

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what about intergral test

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why integral test? we don't have any exponents lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nvm. My final guess is that its similar to the alternating harmonic series, which is absolutely convergent.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but it's conditionally convergent lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sqrt{k}\div x = 1/(k^1/2)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0K^(1/2)/K = 1/(k^(1/2))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah idk this one is a feather

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0>_< watch your language

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for series to converge they must be decreasing, but this one is increasing so that's why it's conditionally convergent ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can use the alternating series test.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I did, but it didn't work >_<

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I ended up with 1 , which means no conclusion, the limit must be equal to 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You just need to show that sqrt(k)/(k+1) goes to zero as k goes to infinity (easy) and that sqrt(k)/(k+1) is monotonic decreasing. You can do this by taking the derivative and showing that the derivative is negative for all x greater than 1. \[f'(x)=\frac{1x}{2\sqrt{x}(x+1)^2}=0 \rightarrow x=1\]partitions the interval. You take a test point x=4, say (anything in this interval) and show that \[f'(4)<0\]. It will be negative for all x in this interval (i.e x >= 1). By the Mean Value Theorem, you can choose any interval in [1, infinity), [a,b], say, and show that since f'(x)<0 on [1,infty), it is the case that\[f'(c)=\frac{f(b)f(a)}{ba}<0 \rightarrow f(b)<f(a)\]i.e. monotonic decreasing.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but the limit is equal to = 1 loki

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{k \rightarrow \infty}\frac{k^{1/2}}{k+1}=\lim_{k \rightarrow \infty}\frac{1/k^{1/2}}{1+1/k}=0\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{k \rightarrow \infty} \frac{\sqrt(k)}{k+1} = \lim_{k \rightarrow \infty} \frac{k}{k} = 1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, it goes to zero...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0didn't I compute it right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I assume your numerator is\[\sqrt{k}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there are alot of ways ^_^ you take the biggest power above, and the biggest power bellow then compute :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=lim+n+goes+to+infty+sqrt%28n%29%2F%281%2Bn%29

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yep, which will be k........oh wait! LOL

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0>_< silly silly mistake, k^1/2  K = 1/k = 0 you're right, I'm sorry ^_^"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so it's limit = 0 and it's decreasing using the ratio test/finding the derivative, so by AST it's convergent?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but the book says it's conditionally convergent?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that's the confusing part

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, the alternating series test is dependent upon the series having that ()^k term in there. If it didn't, the sum wouldn't be killed off enough at every second step.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Absolute convergence would have sqrt(k)/(k+1) as your series, not ()^k etc...different.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm, the theorem says the following: if sigma (ak) is convergent, but sigma ak is divergent, then sigmaak is conditionally convergent. but using AST, both are convergent, which means that ak is absolutely convergent

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0or wait! LOL you can use limit comparison test >_< where : \[ak = \frac{\sqrt(k)}{k+1}\] and\[bk = \frac{1}{\sqrt(k)}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can't use alternating series test on something that isn't alternating.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ak isn't alternating.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol yep! so we can use limit comparison test ! ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then we must find limit of ak/bk

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think you're being crazy. Use the alternating series test.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Truuuuuuuuuuuusssssssssssssssttttttttttttttt mmmmmmmmmmmmmmmeeeeeeeeeeeeeeeeeeeee!!!!!!!!!!!!!!!! :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0>_< lol, but AST gives absolute convergence, there's something wrong

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So are you now saying you want to test for convergence of ak?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when I find the limit of ak/bk = 1 > 0, so it diverges hmmmm

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the question wants you to determine whether the series is absolutely convergent, conditionally convergent or divergent.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait limit comparison test won't work since it says if ak converges, then bk must converge and the vice versa, since they are a couple< as pointed out by my professor lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, if it's not abs. convergent, but convergent otherwise, it's conditionally convergent. If it's not convergent for either, it's divergent. We've shown it's convergent as is, now we show that it is divergent for ak...it will therefore be conditionally convergent.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can use the integral test for the absolute case.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0remember when I said that ak = swrt(k)/k+1 and bk = 1/sqrt(k)? bk is divergent as pseries, 1/2 < 1 so it's divergent, but ak is convergent, so it's conditionally convergent?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why integral test? we use the integral test if we face exponentials and etc

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nah, it won't work, it's either both converge, or both diverge

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hold on, can't we use root test? ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can use limit comparison, with what you have. I got limit =1. It's greater than zero and finite, and since 1/sqrt(k) is divergent, ak is divergent too.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but ak is convergent, so that's why it's conditonally convergent!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOKI, you're a genius LOL

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You're finished. 1) Limit comparison using 1/sqrt{k} to show ak divergent 2) Alternating series to show ak is convergent 3) Therefore, the series is conditionally convergent.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0o_o why use both ways?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you were right just now >_<

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0abt the limit comparison test lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You have to use both ways. Don't type anything until I've finished...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay, I'll behave ._.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Absolute convergence is a stronger deal than conditional convergence. If a series is absolutely convergent, it is conditionally convergent. So abs. conv. is a *sufficient* condition to show convergence. If a series is not absolutely convergent, it doesn't mean it is necessary divergent. It may still be convergent. If it is, and does not have absolute convergence, you have *conditional convergence*. So people attack these problems first by looking for absolute convergence...why? Because the tests are easy and abs. convergence guarantees the other convergence. If it isn't abs. convergent, you STILL have to check for convergence.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so, you're saying that absolute convergence = conditional convergence?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but conditional convergence does not have to equal absolute convergence

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It doesn't go both ways.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that doesn't make sense lol, then why did they put the theorem? why did they say if ak diverges and ak converges, then ak is conditionally convergent?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there must be a reason, since they also said if both converge then it's "absolutely" convergent.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I've just checked my old lecture notes, and it says: Definition (Conditional Convergence): A series \[\sum_{}{}u_n\]which is convergent, but not absolutely convergent, is called *conditionally convergent*.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when is it not absolutely convergent?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0When you show that\[\sum_{}{}a_n\]is divergent.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but we've showed that both are convergent?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, we haven't. It's NOT absolutely convergent using the limit comparison test taking a test series of 1/sqrt{k} (which is divergent).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0O_O okay okay, but that is bk and not ak

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What is bk? I have the following:\[a_k=()^{k+1}\frac{\sqrt{k}}{k+1}\]and \[a_k=\frac{\sqrt{k}}{k+1}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we've let ak = sqrt(k)/k+1 and bk = 1/sqrt(k)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so? .. now we find the limit of ak/bk right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0whic is 0, but since bk isdivergent as pseries 1/2 < 1 then bk is divergent, then what?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0whic is 0, but since bk isdivergent as pseries 1/2 < 1 then bk is divergent, then what?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, and so, since \[\sum_{k=1}^{\infty}\frac{1}{\sqrt{k}}\rightarrow \infty\]and also, since\[\lim_{k \rightarrow \infty}\frac{1/\sqrt{k}}{\left( \frac{\sqrt{k}}{1+k} \right)}=\lim_{k \rightarrow \infty} \frac{k+1}{k} =1\], and as the limit of the ratio is greater than zero and finite, you have by the limit comparison test that the series is divergent.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, but what abt the conditional convergence thing?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay...the proof I just did shows that it is NOT ABSOLUTELY CONVERGENT. So now we ask, is it CONDITIONALLY CONVERGENT? Use the alternating series test to show that \[\sum_{}{}a_k\]is CONVERGENT. Since it's NOT ABSOLUTELY CONVERGENT BUT CONVERGENT in the NONABSOLUTE it is CONDITIONALLY CONVERGENT

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think you've misinterpreted a definition as a theorem.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thank you for being patient with me o_o lol, I can be stubborn sometimes ^_^"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So you're fine with the process?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, I'll just go over it again, thank you ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OK. I'm drawing a summary.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thaaaaaaaaank yoouuuuu looookiiiiiii~

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Did that clear it up?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yep~ :) and I'm showing that to my professor lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0go get some sleep now ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh...well, I hope he agrees!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's 5:30pm here  I'm more inclined to get dinner.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol, go eat then :) and bonne apetite if that's how you spell it lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I assume the spelling's right :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you and your assumptions lol, take care :)
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