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sauravshakya

  • 2 years ago

Can someone PROVE this:

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  1. sauravshakya
    • 2 years ago
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    |dw:1358686236762:dw|

  2. sauravshakya
    • 2 years ago
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    The R.H.S=1/2

  3. sauravshakya
    • 2 years ago
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    It is |dw:1358696599099:dw|

  4. sauravshakya
    • 2 years ago
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    Not (2n-2)*(n-1)

  5. sauravshakya
    • 2 years ago
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    It is (2n-2)C(n-1)

  6. UnkleRhaukus
    • 2 years ago
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    oh, \[^{2n-2}C_{n-1}\]

  7. sauravshakya
    • 2 years ago
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    yep

  8. experimentX
    • 2 years ago
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    |dw:1358757323764:dw|

  9. sauravshakya
    • 2 years ago
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    |dw:1358757562194:dw|

  10. experimentX
    • 2 years ago
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    I think this is untrue, the LHS diverges while RHS converges

  11. sauravshakya
    • 2 years ago
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    How can we say LHS diverges?

  12. experimentX
    • 2 years ago
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    the LHS is of order n, definitely it diverges

  13. nitz
    • 2 years ago
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    RHS is convergent as per p test

  14. experimentX
    • 2 years ago
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    \[ \frac{2 (n-1)^2}{8n} = \frac{n^2 - 2n + 1}{4n} = \left( \frac{n}{4} - \frac{1}{2} \right) + \frac{1}{4n}\] Both part of above diverges.

  15. sauravshakya
    • 2 years ago
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    The actual question is here http://openstudy.com/users/jonask#/updates/50ed3bece4b0d4a537cd13df

  16. nitz
    • 2 years ago
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    LHS is divergent by comparison test

  17. sauravshakya
    • 2 years ago
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    So, it is false?

  18. nitz
    • 2 years ago
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    ya

  19. sauravshakya
    • 2 years ago
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    |dw:1358758646393:dw|

  20. experimentX
    • 2 years ago
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    let me check this sequence with mathematica

  21. sauravshakya
    • 2 years ago
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    what result did u get from mathematica?

  22. experimentX
    • 2 years ago
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    So fat the result is positive for that sequence.

  23. experimentX
    • 2 years ago
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    *far

  24. sauravshakya
    • 2 years ago
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    up to how many terms u tried?

  25. experimentX
    • 2 years ago
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    for 1000 the result is 0.982152

  26. experimentX
    • 2 years ago
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    The compact form of the recurrence relation is \[ a_n = \frac{(2n)!}{2^{2n} (n!)^2 (2n-1)}\]

  27. experimentX
    • 2 years ago
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    Here's a mathematica code for it Table[(2 n)!/((n!)^2 2^(2 n) (2 n - 1)), {n, 1, 10}] and to verify it a = 1/2 For[i = 2, i <= 10, i++, a = (2 i - 3)/(2 i) a; Print[a];] This converges exactly to 1 Sum[Binomial[2 n, n]/(2^(2 n) (2 n - 1)), {n, 1, Infinity}]

  28. experimentX
    • 2 years ago
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    \[ a_n = \frac{(2n)!}{2^{2n} (n!)^2 (2n-1)} = \frac{1}{2^{2n}(2n-1)} \times \binom{2n}{n} \]

  29. youarestupid
    • 2 years ago
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    could somebody give a medal please

  30. sauravshakya
    • 2 years ago
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    medal to whom?

  31. youarestupid
    • 2 years ago
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    me

  32. youarestupid
    • 2 years ago
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    I gave you one

  33. sauravshakya
    • 2 years ago
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    If u will prove it may get more than 1 medal

  34. youarestupid
    • 2 years ago
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    I am only I 6 grade I havent done anything like that before I am sorry. but could you please give me a medal

  35. experimentX
    • 2 years ago
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    you can show that \[ \sum_{n=1}^\infty \frac{1}{2^{2n}(2n-1)} \times \binom{2n}{n} = 1 \] using this http://en.wikipedia.org/wiki/Catalan_number

  36. experimentX
    • 2 years ago
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    http://tinyurl.com/bfj7oep

  37. sauravshakya
    • 2 years ago
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    |dw:1359038723356:dw|But how did u get

  38. experimentX
    • 2 years ago
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    reduce an to a(n-1) ... then a(n-1) to a(n-2) .. until a1

  39. sauravshakya
    • 2 years ago
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    an=(2n-3)/2n a(n-1) an=(2n-3)/2n *(2n-5)/(2n-2) * (2n-7)/(2n-4) *...*3/6 * 1/4 *1/2

  40. sauravshakya
    • 2 years ago
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    then?

  41. experimentX
    • 2 years ago
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    collect 2 first

  42. experimentX
    • 2 years ago
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    (2n-3)(2n-5) ... 5.3.1 ^ there are holes ... fill it (2n-2)(2n-4) ... put the even terms at the top and at the bottom. make 2n! at the top and collect 2 from these at the denominator.

  43. sauravshakya
    • 2 years ago
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    Oh wait...I got it

  44. sauravshakya
    • 2 years ago
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    I did a algebra mistake earlier

  45. sauravshakya
    • 2 years ago
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    Thanx for your help @experimentX

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