Wednesday, 29 June 2016

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




Evaluate lim





I tried it by Stirling's Approximation
n! \approx \sqrt{2\pi n}.n^n e^{-n}
but it leads us to nowhere.



Any hint will be of great help.


Answer



\sin (\pi \sqrt{(n!)^2 -n!} )=\sin (\pi \sqrt{(n!)^2 -n!} -\pi\cdot n! )=\sin \left(\pi \frac{-n!}{\sqrt{(n!)^2 -n!} +n!}\right)=\sin \left(\pi \frac{-1}{\sqrt{1 -\frac{1}{n!}} +1}\right)\to -\sin\frac{\pi}{2} =-1


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