anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\frac{(k+1)\sqrt(k+1)}{k^{3/2} + 2 \sqrt(k)}\] that's where I got stuck after using the ratio test
anonymous
  • anonymous
\[\sum_{k=1}^{\infty}(-1)^{k+1} \frac{\sqrt(k)}{k+1}\] that's the question
anonymous
  • anonymous
oh wow, alas, visitors lol ^_^

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.