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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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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\[\frac{(k+1)\sqrt(k+1)}{k^{3/2} + 2 \sqrt(k)}\] that's where I got stuck after using the ratio test
\[\sum_{k=1}^{\infty}(-1)^{k+1} \frac{\sqrt(k)}{k+1}\] that's the question
oh wow, alas, visitors lol ^_^

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It doesn't say what you are doing? test for convergence?
lol, yes ^_^ but I'm not sure which way to choose, I tried ratio test
umm it looks like you could maybe use divergence test
kth term divergence test? ^_^
so from where I stopped , I can use the k-th term divergence test? right?
hold on
kk ^_^
ok with divergence test you get = 0 so that is inconclusive
so i'll try a ratio
no, so that means it's convergent!
but, the book says it's conditionally convergent . That's where I'm confused :(
oh I see now! The series is increasing, but it's limit tends to zero, so that's why it's conditonally convergent right? ^_^
okay 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
um, that's a good guess. But I don't think so
so with the ratio test I got 1
so again, that means nothing
um so what other tests are there
AST, p-series, root test, geometric series
isn't my logic right lol? the series must be decreasing to be absolutely convergent, but it's increasing instead
I'm not sure enough to say yes or no to that. See what happens when you do more conditionally convergent problems
what about intergral test
OH I think I got it
One sec
why integral test? we don't have any exponents lol
nvm. My final guess is that its similar to the alternating harmonic series, which is absolutely convergent.
but it's conditionally convergent lol
\[\sqrt{k}\div x = 1/(k^1/2)\]
K^(1/2)/K = 1/(k^(1/2))
yeah idk this one is a feather
>_< watch your language
..serious?
lol
for series to converge they must be decreasing, but this one is increasing so that's why it's conditionally convergent ^_^
right? .-.
You can use the alternating series test.
I did, but it didn't work >_<
I ended up with 1 , which means no conclusion, the limit must be equal to 0
You 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. \[f'(x)=\frac{1-x}{2\sqrt{x}(x+1)^2}=0 \rightarrow x=1\]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)}{b-a}<0 \rightarrow f(b)
but the limit is equal to = 1 loki
\[\lim_{k \rightarrow \infty}\frac{k^{1/2}}{k+1}=\lim_{k \rightarrow \infty}\frac{1/k^{1/2}}{1+1/k}=0\]
\[\lim_{k \rightarrow \infty} \frac{\sqrt(k)}{k+1} = \lim_{k \rightarrow \infty} \frac{|k|}{k} = 1\]
No, it goes to zero...
how come? .-.
look above
didn't I compute it right?
I did lol
I assume your numerator is\[\sqrt{k}\]
there are alot of ways ^_^ you take the biggest power above, and the biggest power bellow then compute :)
http://www.wolframalpha.com/input/?i=lim+n+goes+to+infty+sqrt%28n%29%2F%281%2Bn%29
yep, which will be |k|........oh wait! LOL
>_< silly silly mistake, k^1/2 - K = 1/k = 0 you're right, I'm sorry ^_^"
so it's limit = 0 and it's decreasing using the ratio test/finding the derivative, so by AST it's convergent?
Yes
but the book says it's conditionally convergent?
that's the confusing part
Well, 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.
Absolute convergence would have sqrt(k)/(k+1) as your series, not (-)^k etc...different.
Actually, wait...
hmm, the theorem says the following: if sigma (ak) is convergent, but sigma |ak| is divergent, then sigma|ak| is conditionally convergent. but using AST, both are convergent, which means that ak is absolutely convergent
or wait! LOL you can use limit comparison test >_< where : \[ak = \frac{\sqrt(k)}{k+1}\] and\[bk = \frac{1}{\sqrt(k)}\]
You can't use alternating series test on something that isn't alternating.
|ak| isn't alternating.
lol yep! so we can use limit comparison test ! ^_^
then we must find limit of ak/bk
right? :)
I think you're being crazy. Use the alternating series test.
Truuuuuuuuuuuusssssssssssssssttttttttttttttt mmmmmmmmmmmmmmmeeeeeeeeeeeeeeeeeeeee!!!!!!!!!!!!!!!! :)
>_< lol, but AST gives absolute convergence, there's something wrong
So are you now saying you want to test for convergence of |ak|?
when I find the limit of ak/bk = 1 > 0, so it diverges hmmmm
the question wants you to determine whether the series is absolutely convergent, conditionally convergent or divergent.
wait 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
WAIT! I GOT IT
P -SERIES LOL
Okay, 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.
You can use the integral test for the absolute case.
remember when I said that ak = swrt(k)/k+1 and bk = 1/sqrt(k)? bk is divergent as p-series, 1/2 < 1 so it's divergent, but ak is convergent, so it's conditionally convergent?
why integral test? we use the integral test if we face exponentials and etc
nah, it won't work, it's either both converge, or both diverge
have mercy T_T
hold on, can't we use root test? ^_^
You 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.
but ak is convergent, so that's why it's conditonally convergent!
LOKI, you're a genius LOL
You'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.
o_o why use both ways?
you were right just now >_<
abt the limit comparison test lol
You have to use both ways. Don't type anything until I've finished...
okay, I'll behave ._.
Absolute 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.
You can write now :D
so, you're saying that absolute convergence = conditional convergence?
= convergence?
yes
but conditional convergence does not have to equal absolute convergence
It doesn't go both ways.
that 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?
there must be a reason, since they also said if both converge then it's "absolutely" convergent.
I'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*.
when is it not absolutely convergent?
When you show that\[\sum_{}{}|a_n|\]is divergent.
but we've showed that both are convergent?
No, we haven't. It's NOT absolutely convergent using the limit comparison test taking a test series of 1/sqrt{k} (which is divergent).
O_O okay okay, but that is bk and not |ak|
What is bk? I have the following:\[a_k=(-)^{k+1}\frac{\sqrt{k}}{k+1}\]and \[|a_k|=\frac{\sqrt{k}}{k+1}\]
we've let |ak| = sqrt(k)/k+1 and bk = 1/sqrt(k)
so? .-. now we find the limit of ak/bk right?
whic is 0, but since bk isdivergent as p-series 1/2 < 1 then bk is divergent, then what?
whic is 0, but since bk isdivergent as p-series 1/2 < 1 then bk is divergent, then what?
Yes, 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.
yes, but what abt the conditional convergence thing?
Okay...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 NON-ABSOLUTE it is CONDITIONALLY CONVERGENT
O_O.....o.k
phew
I think you've misinterpreted a definition as a theorem.
thank you for being patient with me o_o lol, I can be stubborn sometimes ^_^"
So you're fine with the process?
yes, I'll just go over it again, thank you ^_^
OK. I'm drawing a summary.
alright~
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thaaaaaaaaank yoouuuuu looookiiiiiii~
Did that clear it up?
yep~ :) and I'm showing that to my professor lol
go get some sleep now ^_^
oh...well, I hope he agrees!
It's 5:30pm here - I'm more inclined to get dinner.
he will ~ :)
lol, go eat then :) and bonne apetite if that's how you spell it lol
I assume the spelling's right :)
you and your assumptions lol, take care :)

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