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Jonask

  • 2 years ago

A sequence of numbers (a_n) is defined as

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  1. Jonask
    • 2 years ago
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    \[a_1=1/2\] and for each \[n \ge 2\] \[\huge a_n=\left(\frac{ 2n-3 }{ 2n }\right)a_{n-1}\] Prove that \[\huge\sum_{k=1}^{n}a_k<1 \] for all \[n \ge1\]

  2. sauravshakya
    • 2 years ago
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    |dw:1357738064697:dw|Just try to Prove

  3. Azteck
    • 2 years ago
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    \[a_{1}+ a_{2}+…+ a_{n} < 1\] \[\frac{ 1 }{ 2 } + a_{2} + a_{3}+…+(\frac{ 2n - 3 }{ 2n })a_{n-1} < 1\] For n=1 LHS=\[\frac{ 1 }{ 2 }\] \[<RHS\] (Therefore) True for n=1 For n=2 LHS=\[(\frac{ 2(2) - 3 }{ 2(2) })a_{2-1}\] \[<RHS\] (Therefore) True for n=2

  4. Azteck
    • 2 years ago
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    Assume the formula is true for n=m \[\frac{ 1 }{ 2 } + a_{2} + a_{3}+…+(\frac{ 2m - 3 }{ 2m })a_{m-1} < 1\] For n=m+1 \[\frac{ 1 }{ 2 } + a_{2} + a_{3}+…+(\frac{ 2m - 3 }{ 2m })a_{m-1}+(\frac{ 2(m+1) - 3 }{ 2(m+1) })a_{m+1-1} < 1\] LHS= \[1 + (\frac{ 2(m+1) - 3 }{ 2(m+1) })a_{m}\] \[=(\frac{ 2m-1 }{ 2m+2 })(\frac{ 2m - 3 }{ 2m })a_{m-1}\]

  5. Jonask
    • 2 years ago
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    @sauravshakya @azteck thanks

  6. Jonask
    • 2 years ago
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    not really worth 1 hour

  7. sauravshakya
    • 2 years ago
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    I am not sure if this helps

  8. sauravshakya
    • 2 years ago
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    |dw:1359039410279:dw|

  9. sauravshakya
    • 2 years ago
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    Did u mean like this @experimentX

  10. experimentX
    • 2 years ago
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    yeah .. somewhat similar to that. but one terms is missing.

  11. experimentX
    • 2 years ago
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    there should be one n ... or i made mistake.

  12. sauravshakya
    • 2 years ago
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    |dw:1359040853679:dw|

  13. sauravshakya
    • 2 years ago
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    It was me who did a mistake

  14. experimentX
    • 2 years ago
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    hmm ... now lets use this identity. http://planetmath.org/GeneratingFunctionForTheCatalanNumbers.html

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