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graydarl

i must find lim of a_n for x_n = 1/n( 1/ln(2) + ... + 1/ln(n) ) using stoltz cesaro theorem

  • one year ago
  • one year ago

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  1. mahmit2012
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    xn>int 1/xlnx]1 to inf=ln(lnx)]1 to inf=inf so limxn=inf

    • one year ago
  2. graydarl
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    (1/n)*( 1/ln(2) + ... + 1/ln(n) ) is the equation sorry

    • one year ago
  3. graydarl
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    I must use stoltz cesaro \[\frac{a_{n+1}-a ^{n} }{ b_{n+1}-b ^{n} }\] for \[x _{n}=\frac{ 1 }{ n}*(\frac{ 1 }{ \ln_{2} }+ ... +\frac{ 1 }{ \ln_{n} })\] and i don t know how to find \[a_{n}\] and \[b _{n}\]

    • one year ago
  4. graydarl
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    Please help

    • one year ago
  5. cnknd
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    try b_n = n, and a_n = 1/ln2 + ... + 1/ln(n)

    • one year ago
  6. graydarl
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    b_n is n or 1/n :-S

    • one year ago
  7. graydarl
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    with n get lim = 0 which is good but can i take only n and not 1/n as b_n? :D

    • one year ago
  8. mahmit2012
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    I compare with integral which it approach to infinity then I conclude the series is also infinity or it is diverged.

    • one year ago
  9. graydarl
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    i have a hint and says that the result is 0 but if i consider b_n 1/n the result is infinity so i don't know if the hint is good or not or whick one is correct

    • one year ago
  10. cnknd
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    ugh you want x_n in the form of a_n/b_n... so tell me, what would a_n be if b_n = 1/n?

    • one year ago
  11. graydarl
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    you' re right now i saw that if u write them your way is the same thing, is correct, i understood and the result is 0 as it should be thank you very muc :D i apreciate it

    • one year ago
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