Wednesday, 29 June 2016

algebra precalculus - Evaluate limlimitsntoinftysin2(pisqrt(n!)2(n!))




Evaluate limnsin2(π(n!)2(n!))





I tried it by Stirling's Approximation
n!2πn.nnen


but it leads us to nowhere.



Any hint will be of great help.


Answer



sin(π(n!)2n!)=sin(π(n!)2n!πn!)=sin(πn!(n!)2n!+n!)=sin(π111n!+1)sinπ2=1


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