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JonaskBest ResponseYou've already chosen the best response.0
\[a_1=1/2\] and for each \[n \ge 2\] \[\huge a_n=\left(\frac{ 2n3 }{ 2n }\right)a_{n1}\] Prove that \[\huge\sum_{k=1}^{n}a_k<1 \] for all \[n \ge1\]
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
dw:1357738064697:dwJust try to Prove
 one year ago

AzteckBest ResponseYou've already chosen the best response.1
\[a_{1}+ a_{2}+…+ a_{n} < 1\] \[\frac{ 1 }{ 2 } + a_{2} + a_{3}+…+(\frac{ 2n  3 }{ 2n })a_{n1} < 1\] For n=1 LHS=\[\frac{ 1 }{ 2 }\] \[<RHS\] (Therefore) True for n=1 For n=2 LHS=\[(\frac{ 2(2)  3 }{ 2(2) })a_{21}\] \[<RHS\] (Therefore) True for n=2
 one year ago

AzteckBest ResponseYou've already chosen the best response.1
Assume the formula is true for n=m \[\frac{ 1 }{ 2 } + a_{2} + a_{3}+…+(\frac{ 2m  3 }{ 2m })a_{m1} < 1\] For n=m+1 \[\frac{ 1 }{ 2 } + a_{2} + a_{3}+…+(\frac{ 2m  3 }{ 2m })a_{m1}+(\frac{ 2(m+1)  3 }{ 2(m+1) })a_{m+11} < 1\] LHS= \[1 + (\frac{ 2(m+1)  3 }{ 2(m+1) })a_{m}\] \[=(\frac{ 2m1 }{ 2m+2 })(\frac{ 2m  3 }{ 2m })a_{m1}\]
 one year ago

JonaskBest ResponseYou've already chosen the best response.0
@sauravshakya @azteck thanks
 one year ago

JonaskBest ResponseYou've already chosen the best response.0
not really worth 1 hour
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
I am not sure if this helps
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
dw:1359039410279:dw
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
Did u mean like this @experimentX
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
yeah .. somewhat similar to that. but one terms is missing.
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
there should be one n ... or i made mistake.
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
dw:1359040853679:dw
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
It was me who did a mistake
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
hmm ... now lets use this identity. http://planetmath.org/GeneratingFunctionForTheCatalanNumbers.html
 one year ago
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