## 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?

1. anonymous

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

2. anonymous

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

3. anonymous

oh wow, alas, visitors lol ^_^

4. anonymous

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

5. anonymous

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

6. anonymous

umm it looks like you could maybe use divergence test

7. anonymous

kth term divergence test? ^_^

8. anonymous

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

9. anonymous

hold on

10. anonymous

kk ^_^

11. anonymous

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

12. anonymous

so i'll try a ratio

13. anonymous

no, so that means it's convergent!

14. anonymous

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

15. 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? ^_^

16. 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

17. anonymous

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

18. anonymous

so with the ratio test I got 1

19. anonymous

so again, that means nothing

20. anonymous

um so what other tests are there

21. anonymous

AST, p-series, root test, geometric series

22. anonymous

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

23. anonymous

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

24. anonymous

what about intergral test

25. anonymous

OH I think I got it

26. anonymous

One sec

27. anonymous

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

28. anonymous

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

29. anonymous

but it's conditionally convergent lol

30. anonymous

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

31. anonymous

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

32. anonymous

yeah idk this one is a feather

33. anonymous

>_< watch your language

34. anonymous

..serious?

35. anonymous

lol

36. anonymous

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

37. anonymous

right? .-.

38. anonymous

You can use the alternating series test.

39. anonymous

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

40. anonymous

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

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

42. anonymous

but the limit is equal to = 1 loki

43. anonymous

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

44. anonymous

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

45. anonymous

No, it goes to zero...

46. anonymous

how come? .-.

47. anonymous

look above

48. anonymous

didn't I compute it right?

49. anonymous

I did lol

50. anonymous

I assume your numerator is$\sqrt{k}$

51. anonymous

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

52. anonymous
53. anonymous

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

54. anonymous

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

55. anonymous

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

56. anonymous

Yes

57. anonymous

but the book says it's conditionally convergent?

58. anonymous

that's the confusing part

59. 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.

60. anonymous

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

61. anonymous

Actually, wait...

62. 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

63. anonymous

or wait! LOL you can use limit comparison test >_< where : $ak = \frac{\sqrt(k)}{k+1}$ and$bk = \frac{1}{\sqrt(k)}$

64. anonymous

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

65. anonymous

|ak| isn't alternating.

66. anonymous

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

67. anonymous

then we must find limit of ak/bk

68. anonymous

right? :)

69. anonymous

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

70. anonymous

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

71. anonymous

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

72. anonymous

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

73. anonymous

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

74. anonymous

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

75. 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

76. anonymous

WAIT! I GOT IT

77. anonymous

P -SERIES LOL

78. 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.

79. anonymous

You can use the integral test for the absolute case.

80. 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?

81. anonymous

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

82. anonymous

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

83. anonymous

have mercy T_T

84. anonymous

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

85. 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.

86. anonymous

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

87. anonymous

LOKI, you're a genius LOL

88. 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.

89. anonymous

o_o why use both ways?

90. anonymous

you were right just now >_<

91. anonymous

abt the limit comparison test lol

92. anonymous

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

93. anonymous

okay, I'll behave ._.

94. 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.

95. anonymous

You can write now :D

96. anonymous

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

97. anonymous

= convergence?

98. anonymous

yes

99. anonymous

but conditional convergence does not have to equal absolute convergence

100. anonymous

It doesn't go both ways.

101. 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?

102. anonymous

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

103. 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*.

104. anonymous

when is it not absolutely convergent?

105. anonymous

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

106. anonymous

but we've showed that both are convergent?

107. 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).

108. anonymous

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

109. anonymous

What is bk? I have the following:$a_k=(-)^{k+1}\frac{\sqrt{k}}{k+1}$and $|a_k|=\frac{\sqrt{k}}{k+1}$

110. anonymous

we've let |ak| = sqrt(k)/k+1 and bk = 1/sqrt(k)

111. anonymous

so? .-. now we find the limit of ak/bk right?

112. anonymous

whic is 0, but since bk isdivergent as p-series 1/2 < 1 then bk is divergent, then what?

113. anonymous

whic is 0, but since bk isdivergent as p-series 1/2 < 1 then bk is divergent, then what?

114. 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.

115. anonymous

yes, but what abt the conditional convergence thing?

116. 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

117. anonymous

O_O.....o.k

118. anonymous

phew

119. anonymous

I think you've misinterpreted a definition as a theorem.

120. anonymous

thank you for being patient with me o_o lol, I can be stubborn sometimes ^_^"

121. anonymous

So you're fine with the process?

122. anonymous

yes, I'll just go over it again, thank you ^_^

123. anonymous

OK. I'm drawing a summary.

124. anonymous

alright~

125. anonymous

126. anonymous

thaaaaaaaaank yoouuuuu looookiiiiiii~

127. anonymous

Did that clear it up?

128. anonymous

yep~ :) and I'm showing that to my professor lol

129. anonymous

go get some sleep now ^_^

130. anonymous

oh...well, I hope he agrees!

131. anonymous

It's 5:30pm here - I'm more inclined to get dinner.

132. anonymous

he will ~ :)

133. anonymous

lol, go eat then :) and bonne apetite if that's how you spell it lol

134. anonymous

I assume the spelling's right :)

135. anonymous

you and your assumptions lol, take care :)