## anonymous 5 years ago summation of n=1 to infinty ln((n^2+1)/(2n^2+2)) Does this diverge or converge?

1. anonymous

it converges.

2. anonymous

$\sum_{1}^{\infty} \ln ((n ^{2}+1)/(2n ^{2}+2))$

3. anonymous

how does this converge?

4. anonymous

ln(n^2+1)/(2(n^2+1)) = ln 1/2 is a finite number

5. anonymous

Yes, but for every n you add ln(1/2) And there are a lot of ns...

6. anonymous

sorry i had this one already. my question is rather this: $\sum_{1}^{\infty} \ln ((n ^{2}+1)/(n ^{2}+2))$

7. anonymous

wait, I see you posted the problem again in equation form. Let me see

8. anonymous

yes. the denominator in this case is n squared plus 2

9. anonymous

Newton, I have a doubt, $\sum_{n=1}^{\infty}1/2 = ?$

10. anonymous

with the first one, the limit of the expression was 1/2 and the natural log was not equal to zero so it indeed converged. but i can't determine this one.

11. anonymous

that wouldn't make any sense since there is no n term in the summation. Isn't that right?

12. anonymous

It isn't summing a half once though, it is summing it for every n. n = 1 , add ln(1/2) n = 2 , add ln(1/2) n = .... etc

13. anonymous

okay i understand.

14. anonymous

do you guys have a clue for the question?

15. anonymous

or should repost the question again? if you look above, you'd realize i posted a different but quite similar question.

16. anonymous

for the first problem, it is infinite, for the second problem, as n reaches infinity, the summation reaches 0. So it is finite

17. anonymous

how? that's what i am not getting.

18. anonymous

not the summation, the value of n^2+1/n^2/2

19. anonymous

it's n^2+1/n^2+2

20. anonymous

oh right, my mistake. Let me start afresh.

21. anonymous

i tried the divergence test and it doesn't work. does it converge or diverge?

22. anonymous

$\sum_{n=1}^{\infty}(n ^{2}+1)/(n ^{2}+2)$

23. anonymous

that is the problem correct?

24. anonymous

whoops, its the natural log of that

25. anonymous

yep

26. anonymous

we know that ln 1 =0

27. anonymous

for large values of n, $n ^{2}+1/n ^{2}+2 \approx n ^{2}/n ^{2} =1$

28. anonymous

so as n approaches infinity, the natural log of the given equation approaches 0. so it is a finite number

29. anonymous

therefore it converges

30. anonymous

what i did with the first question was to find the limit which was 1/2 and apply the natural log. With this question, what was your approach?

31. anonymous

In general, the geometric series $\sum_{n=0}^{\infty}z ^{n}$ converges if and only if |z| < 1.

32. anonymous

so you used the absolute convergence test i presume, which i am yet to learn. is that the only way out? can you outline the specifics of your step?

33. anonymous

$\sum_{1}^{\infty} \ln((n^2+1)/(2n^2+2))$ this sum does not converge

34. anonymous

i think you're right since the ln 1/2 is not equal to zero for the divergence test, hence diverges.

35. anonymous

do you see the other question?

36. anonymous

what for example?

37. anonymous

$\sum_{1}^{\infty} \ln ((n ^{2}+1)/(n ^{2}+2))$

38. anonymous

this sum we can find

39. anonymous

how?

40. anonymous

there are two ways to approach this problem. I am not sure which way is correct though. So here goes. $(n ^{2}+1)/(n ^{2}+2) = (n ^{2}+2-1)/(n ^{2}+2)$

41. anonymous

42. anonymous

$=1-1/(n ^{2}+2)$

43. anonymous

yes

44. anonymous

but we have natural logarithm

45. anonymous

for large values of n, 1/n^2+2 approaches 0. so for large values of n, 1- 1/n^2+2 = 1-0 = 1 ln 1 = 0

46. anonymous

no it's not good i think we must do like as...

47. anonymous

yes, ln 1 = 0 even for natural log. log 1 to any base is 0.

48. anonymous

$\sum_{1}^{\infty}\ln(n^2+1)-\sum_{1}^{\infty}\ln(n^2+2)$

49. anonymous

you guys have me confused now. which approach is agreed upon now. and does it converge or not? can you also state the test you are using?

50. anonymous

it converges. That is agreed upon. The method is what we are disagreeing upon.

51. anonymous

what i have learnt so far uses the various test methods like divergence test, integral test, basic comparison test, etc to solve this questions. which one are you applying in this case?

52. anonymous

but this sums does not converge =)

53. anonymous

are you talking about his sum or the previous one, hobbit?

54. anonymous

because as far as I can tell, this sum seems to be converging.

55. anonymous

but this sums does not converge =)

56. anonymous

sorry i have some problems with charge on my mac)))

57. anonymous

oh well, then you should explain it to bbcram. because I seem to be wrong on this one.

58. anonymous

what do you don't understand bbcram?

59. anonymous

so guys i have a question)) where you from?

60. anonymous

im from india Please explain how you got this sum is diverging.

61. anonymous

cool i am from Russia you must to know that ln(a/b)=ln(a)-ln(b)

62. anonymous

Russian eh? Cool. Student? yes ln(a/b)=lna-lnb

63. anonymous

and $\sum_{?}^{?}(a-b)=\sum_{?}^{?}a-\sum_{?}^{?}b$

64. anonymous

Does Hobbit think: $\sum_{1}^{\infty} \ln \frac{n^2+1}{n^2+2}$ DIVERGES? xD

65. anonymous

yes

66. anonymous

yes, he seems to think so. I am asking him why he thinks that

67. anonymous

68. anonymous

huh?

69. anonymous

don't worry Hobbit, Wolfram |Alpha thinks so too ... xD

70. anonymous

That's not divergence, kiddo.

71. anonymous

it s partial sum

72. anonymous

INewton where you from? and where do you study?

73. anonymous

sorry guys my pc suddenly restarted. It's no less clear now. INewton, dhatraditya, Hobbit. I am in the united states. I just started the topic on sequence and series this week and were challenged with this question. I have worked on others successfully but this is a headache. So what is your take? I seem to be seeing conflicting answers and approaches

74. anonymous
75. anonymous

that is a finite number. So it is convergent.

76. anonymous

yes

77. anonymous

Then why did you think it was divergent?

78. anonymous

i dont think that it is divergent

79. anonymous

you said it does not converge.

80. anonymous

∑1∞ln(n2+1) is divergent and ∑1∞ln(n2+2) is divergent

81. anonymous

is not it?

82. anonymous

yes.

83. anonymous

ok, so the ln of either the numerator or denominator in this case diverges. cool. however, the ln for the equation is 0 right? which makes it convergent?

84. anonymous

i don't know, but may be gamma function ?

85. anonymous

i in confuse too now =)

86. anonymous

87. anonymous

@dhatradtiya, can you just once again write the steps you took so i can write that down and study your procedure?

88. anonymous

yes, $(n ^{2}+1)/(n ^{2}+2)=((n ^{2}+2)−1))/(n ^{2}+2)$

89. anonymous

$=1−1/(n ^{2}+2)$

90. anonymous

for large values of n,

91. anonymous

$1/(n ^{2}+2) \approx0$

92. anonymous

yes and so

93. anonymous

therefore $1−1/(n ^{2}+2) \approx1-0 = 1$

94. anonymous

but 1-1=0 not 1

95. anonymous

$n \rightarrow \infty, \ln (1-1/(n ^{2}+2) = 0$

96. anonymous

for n=1 ln(1-1/3) for n=2 ln(1-1/6) for n=3 ln(1-1/11) for n=infty ln(1-0)=0

97. anonymous

yes but we must to find sum of first members

98. anonymous

ln(1-1/3)+ln(1-1/6)+ln(1-1/11)+...

99. anonymous

It has been quite a while since I did my last convergence test. It seems obvious to me, but for proof, you will need to ask someone else. I can only give you my reasoning as to why it is convergent.

100. anonymous

thanks guys. are you mostly available on this forum?

101. anonymous

I come here in my free time often to help out.

102. anonymous

cool. can i give you my email or you give me yours. i could keep in touch

103. anonymous

i'm in university now =)

104. anonymous

did you understand the reasoning, bbcram? It is not important to know the exact values of the series. You should be able to look at a series and tell if it is convergent or not directly.

105. anonymous

ok cool what is you name? my e-mail: galexant@yandex.ru

106. anonymous

107. anonymous

108. anonymous

i'm in the university too, except that with my calculus 11 professor essentially leaves everything to us. i would look over it carefully to understand the reasoning better and make good deductions.

109. anonymous

cool. mine is rankbert@gmail.com. i use this more often.

110. anonymous

dhatraditya you right thanks for discussion in discussion birth truth

111. anonymous

yes, it was a great discussion. Good luck to both of you!

112. anonymous

i like the way it all went. you guys are cool. thanks and stay in touch. i would stay in touch too.

113. anonymous

thanks good luck to you too

114. anonymous

goodluck to you both too

115. anonymous

yes i would stay in touch

116. anonymous

@bbcram in what university do you study

117. anonymous

CUNY in New York

118. anonymous

and what is you specialization?

119. anonymous

computer science. We can have further conversation through email if you don't mind. Thanks for your time and dhatraditya too. i'm out

120. anonymous

ok ciao guys.

121. anonymous

cool i am study on computer science too

122. anonymous

ciao

123. anonymous

see soon here