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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    it converges.

  2. anonymous
    • 5 years ago
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    \[\sum_{1}^{\infty} \ln ((n ^{2}+1)/(2n ^{2}+2))\]

  3. anonymous
    • 5 years ago
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    how does this converge?

  4. anonymous
    • 5 years ago
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    ln(n^2+1)/(2(n^2+1)) = ln 1/2 is a finite number

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

  6. anonymous
    • 5 years ago
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    sorry i had this one already. my question is rather this: \[\sum_{1}^{\infty} \ln ((n ^{2}+1)/(n ^{2}+2))\]

  7. anonymous
    • 5 years ago
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    wait, I see you posted the problem again in equation form. Let me see

  8. anonymous
    • 5 years ago
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    yes. the denominator in this case is n squared plus 2

  9. anonymous
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    Newton, I have a doubt, \[\sum_{n=1}^{\infty}1/2 = ?\]

  10. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    that wouldn't make any sense since there is no n term in the summation. Isn't that right?

  12. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    okay i understand.

  14. anonymous
    • 5 years ago
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    do you guys have a clue for the question?

  15. anonymous
    • 5 years ago
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    or should repost the question again? if you look above, you'd realize i posted a different but quite similar question.

  16. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    how? that's what i am not getting.

  18. anonymous
    • 5 years ago
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    not the summation, the value of n^2+1/n^2/2

  19. anonymous
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    it's n^2+1/n^2+2

  20. anonymous
    • 5 years ago
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    oh right, my mistake. Let me start afresh.

  21. anonymous
    • 5 years ago
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    i tried the divergence test and it doesn't work. does it converge or diverge?

  22. anonymous
    • 5 years ago
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    \[\sum_{n=1}^{\infty}(n ^{2}+1)/(n ^{2}+2)\]

  23. anonymous
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    that is the problem correct?

  24. anonymous
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    whoops, its the natural log of that

  25. anonymous
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    yep

  26. anonymous
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    we know that ln 1 =0

  27. anonymous
    • 5 years ago
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    for large values of n, \[n ^{2}+1/n ^{2}+2 \approx n ^{2}/n ^{2} =1\]

  28. anonymous
    • 5 years ago
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    so as n approaches infinity, the natural log of the given equation approaches 0. so it is a finite number

  29. anonymous
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    therefore it converges

  30. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    In general, the geometric series \[\sum_{n=0}^{\infty}z ^{n}\] converges if and only if |z| < 1.

  32. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    \[\sum_{1}^{\infty} \ln((n^2+1)/(2n^2+2))\] this sum does not converge

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

  35. anonymous
    • 5 years ago
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    do you see the other question?

  36. anonymous
    • 5 years ago
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    what for example?

  37. anonymous
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    \[\sum_{1}^{\infty} \ln ((n ^{2}+1)/(n ^{2}+2))\]

  38. anonymous
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    this sum we can find

  39. anonymous
    • 5 years ago
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    how?

  40. anonymous
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    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
    • 5 years ago
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    dhatraditya you right

  42. anonymous
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    \[=1-1/(n ^{2}+2)\]

  43. anonymous
    • 5 years ago
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    yes

  44. anonymous
    • 5 years ago
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    but we have natural logarithm

  45. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    no it's not good i think we must do like as...

  47. anonymous
    • 5 years ago
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    yes, ln 1 = 0 even for natural log. log 1 to any base is 0.

  48. anonymous
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    \[\sum_{1}^{\infty}\ln(n^2+1)-\sum_{1}^{\infty}\ln(n^2+2)\]

  49. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    it converges. That is agreed upon. The method is what we are disagreeing upon.

  51. anonymous
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    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
    • 5 years ago
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    but this sums does not converge =)

  53. anonymous
    • 5 years ago
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    are you talking about his sum or the previous one, hobbit?

  54. anonymous
    • 5 years ago
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    because as far as I can tell, this sum seems to be converging.

  55. anonymous
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    but this sums does not converge =)

  56. anonymous
    • 5 years ago
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    sorry i have some problems with charge on my mac)))

  57. anonymous
    • 5 years ago
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    oh well, then you should explain it to bbcram. because I seem to be wrong on this one.

  58. anonymous
    • 5 years ago
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    what do you don't understand bbcram?

  59. anonymous
    • 5 years ago
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    so guys i have a question)) where you from?

  60. anonymous
    • 5 years ago
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    im from india Please explain how you got this sum is diverging.

  61. anonymous
    • 5 years ago
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    cool i am from Russia you must to know that ln(a/b)=ln(a)-ln(b)

  62. anonymous
    • 5 years ago
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    Russian eh? Cool. Student? yes ln(a/b)=lna-lnb

  63. anonymous
    • 5 years ago
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    and \[\sum_{?}^{?}(a-b)=\sum_{?}^{?}a-\sum_{?}^{?}b\]

  64. anonymous
    • 5 years ago
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    Does Hobbit think: \[\sum_{1}^{\infty} \ln \frac{n^2+1}{n^2+2}\] DIVERGES? xD

  65. anonymous
    • 5 years ago
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    yes

  66. anonymous
    • 5 years ago
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    yes, he seems to think so. I am asking him why he thinks that

  67. anonymous
    • 5 years ago
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    answer near -0.956448

  68. anonymous
    • 5 years ago
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    huh?

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

  70. anonymous
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    That's not divergence, kiddo.

  71. anonymous
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    it s partial sum

  72. anonymous
    • 5 years ago
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    INewton where you from? and where do you study?

  73. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    that is a finite number. So it is convergent.

  75. anonymous
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    yes

  76. anonymous
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    Then why did you think it was divergent?

  77. anonymous
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    i dont think that it is divergent

  78. anonymous
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    you said it does not converge.

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

  80. anonymous
    • 5 years ago
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    is not it?

  81. anonymous
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    yes.

  82. anonymous
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    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?

  83. anonymous
    • 5 years ago
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    i don't know, but may be gamma function ?

  84. anonymous
    • 5 years ago
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    i in confuse too now =)

  85. anonymous
    • 5 years ago
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    i think dhatraditya is right.

  86. anonymous
    • 5 years ago
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    @dhatradtiya, can you just once again write the steps you took so i can write that down and study your procedure?

  87. anonymous
    • 5 years ago
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    yes, \[(n ^{2}+1)/(n ^{2}+2)=((n ^{2}+2)−1))/(n ^{2}+2)\]

  88. anonymous
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    \[=1−1/(n ^{2}+2)\]

  89. anonymous
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    for large values of n,

  90. anonymous
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    \[1/(n ^{2}+2) \approx0\]

  91. anonymous
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    yes and so

  92. anonymous
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    therefore \[1−1/(n ^{2}+2) \approx1-0 = 1\]

  93. anonymous
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    but 1-1=0 not 1

  94. anonymous
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    \[n \rightarrow \infty, \ln (1-1/(n ^{2}+2) = 0\]

  95. anonymous
    • 5 years ago
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    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

  96. anonymous
    • 5 years ago
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    yes but we must to find sum of first members

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

  98. anonymous
    • 5 years ago
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    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.

  99. anonymous
    • 5 years ago
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    thanks guys. are you mostly available on this forum?

  100. anonymous
    • 5 years ago
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    I come here in my free time often to help out.

  101. anonymous
    • 5 years ago
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    cool. can i give you my email or you give me yours. i could keep in touch

  102. anonymous
    • 5 years ago
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    i'm in university now =)

  103. anonymous
    • 5 years ago
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    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.

  104. anonymous
    • 5 years ago
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    ok cool what is you name? my e-mail: galexant@yandex.ru

  105. anonymous
    • 5 years ago
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    my email id is dhatraditya@yahoo.com

  106. anonymous
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    what is your name bbcram?

  107. anonymous
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    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.

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

  109. anonymous
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    dhatraditya you right thanks for discussion in discussion birth truth

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

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

  112. anonymous
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    thanks good luck to you too

  113. anonymous
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    goodluck to you both too

  114. anonymous
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    yes i would stay in touch

  115. anonymous
    • 5 years ago
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    @bbcram in what university do you study

  116. anonymous
    • 5 years ago
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    CUNY in New York

  117. anonymous
    • 5 years ago
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    and what is you specialization?

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

  119. anonymous
    • 5 years ago
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    ok ciao guys.

  120. anonymous
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    cool i am study on computer science too

  121. anonymous
    • 5 years ago
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    ciao

  122. anonymous
    • 5 years ago
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    see soon here

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