Question

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?

Answer 1

\[\frac{(k+1)\sqrt(k+1)}{k^{3/2} + 2 \sqrt(k)}\]
that's where I got stuck after using the ratio test

Answer 2

\[\sum_{k=1}^{\infty}(-1)^{k+1} \frac{\sqrt(k)}{k+1}\]
that's the question

Answer 3

oh wow, alas, visitors lol ^_^

Answer 4

It doesn't say what you are doing? test for convergence?

Answer 5

lol, yes ^_^ but I'm not sure which way to choose, I tried ratio test

Answer 6

umm it looks like you could maybe use divergence test

Answer 7

kth term divergence test? ^_^

Answer 8

so from where I stopped , I can use the k-th term divergence test? right?

Answer 9

hold on

Answer 10

kk ^_^

Answer 11

ok with divergence test you get = 0 so that is inconclusive

Answer 12

so i'll try a ratio

Answer 13

no, so that means it's convergent!

Answer 14

but, the book says it's conditionally convergent . That's where I'm confused :(

Answer 15

oh I see now! The series is increasing, but it's limit tends to zero, so that's why it's conditonally convergent right? ^_^

Answer 16

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

Answer 17

um, that's a good guess. But I don't think so

Answer 18

so with the ratio test I got 1

Answer 19

so again, that means nothing

Answer 20

um so what other tests are there

Answer 21

AST, p-series, root test, geometric series

Answer 22

isn't my logic right lol? the series must be decreasing to be absolutely convergent, but it's increasing instead

Answer 23

I'm not sure enough to say yes or no to that. See what happens when you do more conditionally convergent problems

Answer 24

what about intergral test

Answer 25

OH I think I got it

Answer 26

One sec

Answer 27

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

Answer 28

nvm. My final guess is that its similar to the alternating harmonic series, which is absolutely convergent.

Answer 29

but it's conditionally convergent lol

Answer 30

\[\sqrt{k}\div x = 1/(k^1/2)\]

Answer 31

K^(1/2)/K = 1/(k^(1/2))

Answer 32

yeah idk this one is a feather

Answer 33

>_< watch your language

Answer 34

..serious?

Answer 35

lol

Answer 36

for series to converge they must be decreasing, but this one is increasing so that's why it's conditionally convergent ^_^

Answer 37

right? .-.

Answer 38

You can use the alternating series test.

Answer 39

I did, but it didn't work >_<

Answer 40

I ended up with 1 , which means no conclusion, the limit must be equal to 0

Answer 41

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\[f'(c)=\frac{f(b)-f(a)}{b-a}<0 \rightarrow f(b)<f(a)\]i.e. monotonic decreasing.

Answer 42

but the limit is equal to = 1 loki

Answer 43

\[\lim_{k \rightarrow \infty}\frac{k^{1/2}}{k+1}=\lim_{k \rightarrow \infty}\frac{1/k^{1/2}}{1+1/k}=0\]

Answer 44

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

Answer 45

No, it goes to zero...

Answer 46

how come? .-.

Answer 47

look above

Answer 48

didn't I compute it right?

Answer 49

I did lol

Answer 50

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

Answer 51

there are alot of ways ^_^ you take the biggest power above, and the biggest power bellow then compute :)

Answer 52

http://www.wolframalpha.com/input/?i=lim+n+goes+to+infty+sqrt%28n%29%2F%281%2Bn%29

Answer 53

yep, which will be |k|........oh wait! LOL

Answer 54

>_< silly silly mistake, k^1/2 - K = 1/k = 0 you're right, I'm sorry ^_^"

Answer 55

so it's limit = 0 and it's decreasing using the ratio test/finding the derivative, so by AST it's convergent?

Answer 56

Yes

Answer 57

but the book says it's conditionally convergent?

Answer 58

that's the confusing part

Answer 59

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.

Answer 60

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

Answer 61

Actually, wait...

Answer 62

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

Answer 63

or wait! LOL you can use limit comparison test >_< where :

\[ak = \frac{\sqrt(k)}{k+1}\] and\[bk = \frac{1}{\sqrt(k)}\]

Answer 64

You can't use alternating series test on something that isn't alternating.

Answer 65

|ak| isn't alternating.

Answer 66

lol yep! so we can use limit comparison test ! ^_^

Answer 67

then we must find limit of ak/bk

Answer 68

right? :)

Answer 69

I think you're being crazy. Use the alternating series test.

Answer 70

Truuuuuuuuuuuusssssssssssssssttttttttttttttt mmmmmmmmmmmmmmmeeeeeeeeeeeeeeeeeeeee!!!!!!!!!!!!!!!! :)

Answer 71

>_< lol, but AST gives absolute convergence, there's something wrong

Answer 72

So are you now saying you want to test for convergence of |ak|?

Answer 73

when I find the limit of ak/bk = 1 > 0, so it diverges hmmmm

Answer 74

the question wants you to determine whether the series is absolutely convergent, conditionally convergent or divergent.

Answer 75

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

Answer 76

WAIT! I GOT IT

Answer 77

P -SERIES LOL

Answer 78

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.

Answer 79

You can use the integral test for the absolute case.

Answer 80

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?

Answer 81

why integral test? we use the integral test if we face exponentials and etc

Answer 82

nah, it won't work, it's either both converge, or both diverge

Answer 83

have mercy T_T

Answer 84

hold on, can't we use root test? ^_^

Answer 85

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.

Answer 86

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

Answer 87

LOKI, you're a genius LOL

Answer 88

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.

Answer 89

o_o why use both ways?

Answer 90

you were right just now >_<

Answer 91

abt the limit comparison test lol

Answer 92

You have to use both ways. Don't type anything until I've finished...

Answer 93

okay, I'll behave ._.

Answer 94

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.

Answer 95

You can write now :D

Answer 96

so, you're saying that absolute convergence = conditional convergence?

Answer 97

= convergence?

Answer 98

yes