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imqwerty

  • one year ago

.......question :)

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  1. imqwerty
    • one year ago
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    \[\sum_{n=1}^{\infty}\left( \frac{ \sqrt{3} }{ 2n } \right)^2 = \left[ \sum_{n=1}^{\infty}\left( \frac{ 1 }{ mn+l} \right) \right]^2\] m,l are integers :) Find m and l

  2. inkyvoyd
    • one year ago
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    sir are you sure that the numerator of your second expression isn't (-1)^n? I iz confuzzeld

  3. imqwerty
    • one year ago
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    yes i am sure :)

  4. anonymous
    • one year ago
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    will check later

  5. anonymous
    • one year ago
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    \(\Large \sum_{n=1}^{\infty}\left( \frac{ \sqrt{3} }{ 2n } \right)^2 = 3/4 \sum_{n=1}^{\infty} \frac{ {1} }{ n^2 } =\frac{ {3} }{ 4}.\frac{ {\pi^2} }{ 6}=\frac{ {\pi^2} }{ 2}\)

  6. anonymous
    • one year ago
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    \(\Large \left[ \sum_{n=1}^{\infty}\left( \frac{ 1 }{ mn+l} \right) \right]^2 =\dfrac{\pi^2}{2} \\\Large \sum_{n=1}^{\infty}\left( \frac{ 1 }{ mn+l} \right) =\dfrac{\pi^4}{4} \)

  7. anonymous
    • one year ago
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    confused as the first series sounds harmonic series @ganeshie8

  8. imqwerty
    • one year ago
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    yes..and we don have any summation formula for harmonic series :( there must be some trick

  9. anonymous
    • one year ago
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    ok let me think

  10. imqwerty
    • one year ago
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    nd that shuld be pi/sqrt{2}

  11. anonymous
    • one year ago
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    why so ?

  12. imqwerty
    • one year ago
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    http://prntscr.com/8ftthq

  13. anonymous
    • one year ago
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    lol yes sorry :3

  14. ganeshie8
    • one year ago
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    Clearly there is something wrong with the question. The harmonic series doesn't converge

  15. mathmate
    • one year ago
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    Yeah, the first one is \(\pi^2/8\), and the second one is supposed to diverge whatever finite value we put on l and m. I suspect (as @inkyvoyd did) it's an altenating series, in which case we can get the Leibniz's series when m=1 and l=0 to get \(\pi/4\). Then a few adjustments should give \(\pi^2/8\).

  16. anonymous
    • one year ago
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    :P yeah for sure

  17. anonymous
    • one year ago
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    can we at least share songs lol xD

  18. anonymous
    • one year ago
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    https://www.youtube.com/watch?v=mWRsgZuwf_8&list=PLPN1tL8FFbmNz7sWLYNyc_D8YFjiVNU3c&index=56

  19. imqwerty
    • one year ago
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    hey wait there was a typo

  20. anonymous
    • one year ago
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    lol ok what the typo is ?

  21. mathmate
    • one year ago
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    \(\sum_{n=1}^{\infty}\left( \frac{ \sqrt{3} }{ 2n } \right)^2 = \left[ \sum_{n=1}^{\infty}\left( \frac{ (-1)^n }{ mn+l} \right) \right]^2\) right?

  22. imqwerty
    • one year ago
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    it was meant to be this- \[\sum_{n=1}^{\infty}\left( \frac{ \sqrt{3} }{ 2n } \right)=\sum_{n=1}^{\infty}\left( \frac{ 1 }{ mn+l } \right)^2\]

  23. anonymous
    • one year ago
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    -.- same problem see sqrt3/2n is also harmonic

  24. imqwerty
    • one year ago
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    but i know the solution now :D

  25. anonymous
    • one year ago
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    really ?

  26. anonymous
    • one year ago
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    share

  27. imqwerty
    • one year ago
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    ok jst a min :)

  28. imqwerty
    • one year ago
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    on the LHS u have -\[\sum_{n=1}^{\infty}\frac{ 3 }{ 4n}\]take 3/4 out and write it as 1-1/4 nd u get -\[\left( 1-\frac{ 1 }{ 4} \right)\left( \sum_{n=1}^{\infty}\frac{ 1 }{ n^2 }\right)\]open the brackets nd get this-\[\sum_{n=1}^{\infty }\frac{ 1 }{ n^2 } - \sum_{n=1}^{\infty}\frac{ 1 }{ (2n)^2 }\]opening the terms u get this kind of series - \[1-\frac{ 1 }{ 2^2 }+\frac{ 1 }{ 2^2 } -\frac{ 1 }{ 4^2 }+\frac{ 1 }{ 4^2 }........\]+ u get the odd denominator terms too..so finally after cancellations u get this on the LHS- \[\sum_{n=1}^{\infty}\frac{ 1 }{ (2n-1)^2 } \] which u can write as- \[\sum_{n=1}^{\infty}\left( \frac{ 1 }{ 2n-1 } \right)^2 \] so comparing it to the RHS u get m=2 and l= -1

  29. anonymous
    • one year ago
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    :)

  30. imqwerty
    • one year ago
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    hehe (:

  31. mathmate
    • one year ago
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    From what I can see from the solution, the left hand side was meant to be: \((\sum_{n=1}^{\infty}\left( \frac{ \sqrt{3} }{ 2n } \right))^2=\sum_{n=1}^{\infty}\left( \frac{ 1 }{ mn+l } \right)^2\)

  32. imqwerty
    • one year ago
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    yes i jst opened up that square nd wrote the thing initially :)

  33. imqwerty
    • one year ago
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    yes @mathmate that shuld be squared sry for so many typos

  34. mathmate
    • one year ago
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    That's ok. Too bad LaTex didn't make it much easier!

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