anonymous
  • anonymous
A sequence of numbers (a_n) is defined as
Precalculus
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[a_1=1/2\] 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 \[n \ge1\]
anonymous
  • anonymous
|dw:1357738064697:dw|Just try to Prove
anonymous
  • anonymous
\[a_{1}+ a_{2}+…+ a_{n} < 1\] \[\frac{ 1 }{ 2 } + a_{2} + a_{3}+…+(\frac{ 2n - 3 }{ 2n })a_{n-1} < 1\] For n=1 LHS=\[\frac{ 1 }{ 2 }\] \[

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anonymous
  • anonymous
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= \[1 + (\frac{ 2(m+1) - 3 }{ 2(m+1) })a_{m}\] \[=(\frac{ 2m-1 }{ 2m+2 })(\frac{ 2m - 3 }{ 2m })a_{m-1}\]
anonymous
  • anonymous
@sauravshakya @azteck thanks
anonymous
  • anonymous
not really worth 1 hour
anonymous
  • anonymous
I am not sure if this helps
anonymous
  • anonymous
|dw:1359039410279:dw|
anonymous
  • anonymous
Did u mean like this @experimentX
experimentX
  • experimentX
yeah .. somewhat similar to that. but one terms is missing.
experimentX
  • experimentX
there should be one n ... or i made mistake.
anonymous
  • anonymous
|dw:1359040853679:dw|
anonymous
  • anonymous
It was me who did a mistake
experimentX
  • experimentX
hmm ... now lets use this identity. http://planetmath.org/GeneratingFunctionForTheCatalanNumbers.html

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