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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 ^_^

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 :(

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

what about intergral test

OH I think I got it

One sec

why integral test? we don't have any exponents lol

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

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

\[\lim_{k \rightarrow \infty} \frac{\sqrt(k)}{k+1} = \lim_{k \rightarrow \infty} \frac{|k|}{k} = 1\]

No, it goes to zero...

how come? .-.

look above

didn't I compute it right?

I did lol

I assume your numerator is\[\sqrt{k}\]

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 ^_^"

Yes

but the book says it's conditionally convergent?

that's the confusing part

Absolute convergence would have sqrt(k)/(k+1) as your series, not (-)^k etc...different.

Actually, wait...

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

WAIT! I GOT IT

P -SERIES LOL

You can use the integral test for the absolute case.

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? ^_^

but ak is convergent, so that's why it's conditonally convergent!

LOKI, you're a genius LOL

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 ._.

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.

there must be a reason, since they also said if both converge then it's "absolutely" convergent.

when is it not absolutely convergent?

When you show that\[\sum_{}{}|a_n|\]is divergent.

but we've showed that both are convergent?

O_O okay okay, but that is bk and not |ak|

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, but what abt the conditional convergence thing?

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~

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 :)