Wednesday, 1 January 2014

real analysis - Prove that the sequence an=frac1cdot3cdot5cdots(2n1)2cdot4cdot6cdots(2n) is monotonically decreasing sequence




Prove that the sequence an=135(2n1)246(2n)

is monotonically decreasing sequence. I tried an+1an<0, but i was not able to do it.



Help will be appreciated.



Thanks!


Answer



Notice, we have an=1.3.5(2n1)2.4.6(2n)


=[1.3.5(2n1)][2.4.6(2n)][2.4.6(2n)][(2.4.6(2n)]
=1.2.3.4.5(2n1)(2n)[2.4.6(2n)]2
=(2n)![2n(1.2.3(n))]2

=(2n)![2n(n!)]2

an=(2n)!22n(n!)2
an+1=(2n+2)!22n+2((n+1)!)2
an+1=(2n+2)(2n+1)(2n)!422n((n+1)n!)2


=2(n+1)(2n+1)(2n)!4(n+1)222n(n!)2

an+1=(2n+1)(2n)!2(n+1)22n(n!)2



Now, dividing (2) by (1), we get an+1an=(2n+1)(2n)!2(n+1)22n(n!)2(2n)!22n(n!)2

=2n+12(n+1)
an+1an=2n+12n+2<1


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