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Math Help - sums with lnx

  1. #1
    Member akhayoon's Avatar
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    sums with lnx

    I'm having trouble trying to figure out what to do for convergence and divergence for a series that has lnx in it

    for example

    \sum\frac{lnk}{e^{k}} and \sum\frac{lnk}{k^{2}}
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by akhayoon View Post
    I'm having trouble trying to figure out what to do for convergence and divergence for a series that has lnx in it

    for example

    \sum\frac{lnk}{e^{k}} and \sum\frac{lnk}{k^{2}}
    try the integral test
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  3. #3
    Moo
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    Hello,

    I think ratio test can work too...
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Jhevon View Post
    try the integral test
    ...the first one would yield an Ei(x) function...........................thats gross.........or am I integrating wrong? ....
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Moo View Post
    Hello,

    I think ratio test can work too...
    the ratio test is inconclusive

    Quote Originally Posted by Mathstud28 View Post
    ...the first one would yield an Ei(x) function...........................thats gross.........or am I integrating wrong? ....
    you can use the integral test for the second and then use the comparison test for the first in comparing the second with the first
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Jhevon View Post
    the ratio test is inconclusive

    you can use the integral test for the second and then use the comparison test for the first in comparing the second with the first
    ...ahh...I thoguth you just meant to do the integral test....
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    ...ahh...I thoguth you just meant to do the integral test....
    yes, my apologies. I am a man of few words (except in MHF chat ) and so it is easy to misunderstand me....
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  8. #8
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    \ln (n) \leqslant 2\sqrt n  \Rightarrow \quad \frac{{\ln (n)}}<br />
{{n^2 }} \leqslant \frac{2}<br />
{{n^{3/2} }}
    Simple comparison with p-series.
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  9. #9
    Member akhayoon's Avatar
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    I don't understand how that comparison works...I mean where did the 2\sqrt{n} come from and how did it become to the exponent 3/2?? so confused
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by akhayoon View Post
    I don't understand how that comparison works...I mean where did the 2\sqrt{n} come from and how did it become to the exponent 3/2?? so confused
    He just picked a term that was greater than ln(x)...which is 2\sqrt{x}...and he got that from the fact that \frac{\sqrt{x}}{x^2}=x^{-2+\frac{1}{2}}=x^{\frac{-3}{2}}
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  11. #11
    Member akhayoon's Avatar
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    so that was completely arbitrary and u could've essentially picked anything greater than lnx??
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  12. #12
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    He just picked a term that was greater than ln(x)...which is 2\sqrt{x}...and he got that from the fact that \frac{\sqrt{x}}{x^2}=x^{-2+\frac{1}{2}}=x^{\frac{-3}{2}}
    i think the posters point is, how did he know what to pick?
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  13. #13
    Member akhayoon's Avatar
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    well how would you even integrate \int \frac{lnx}{e^{x}}

    by parts?
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  14. #14
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by akhayoon View Post
    well how would you even integrate \int \frac{lnx}{e^{x}}

    by parts?
    You can't do it..well not by normal means...by power series you coud...but to be blun \int\frac{\ln(x)}{e^{x}}dx=\frac{-\ln(x)}{e^{x}}-Ei_1(x) where Ei(x) is the exponential integral is E_1(x)=\int_x^{\infty}\frac{e^{-t}}{t}dt
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  15. #15
    Member akhayoon's Avatar
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    \sum\frac{lnk}{e^{k}}

    so the book is telling me to use the ratio test...

    and you get \frac{ln(k+1)}{elnk}=\frac{1}{e}

    so now I guess how would ln(k+1)/lnk go to 1 would be my basic question...

    all I know is is that ln(n)/n goes to 0 as n goes to infinity but i haven't seem something like this before..
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