sum n=o to oo (2n)!(x/2)^n = (2n+2)!(x/2)^(n+1) / (2n)!(x/2)^n = (2n+2)(2n+1)(x/2) where do i go from there? ><
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Originally Posted by jeph sum n=o to oo (2n)!(x/2)^n = (2n+2)!(x/2)^(n+1) / (2n)!(x/2)^n = (2n+2)(2n+1)(x/2) where do i go from there? >< take the limit as n goes to infinity, and set whatever the result is to be < 1. then you find the x's that satisfy it and check the end points
Originally Posted by jeph sum n=o to oo (2n)!(x/2)^n = (2n+2)!(x/2)^(n+1) / (2n)!(x/2)^n = (2n+2)(2n+1)(x/2) where do i go from there? >< Note, Ratio test for , So, this convergense only for
so does it go lim n to oo [(2n+2)(2n+1)/2]x = |[oo/2]x|<1 -1< [oo/2]x < 1 -2/oo< x < 2/oo 0< x <0 x=0 ?
Originally Posted by jeph so does it go lim n to oo [(2n+2)(2n+1)/2]x = |[oo/2]x|<1 -1< [oo/2]x < 1 -2/oo< x < 2/oo 0< x <0 x=0 ? Is there a number such that ? No! But the thing is that we assumed (secretly) that . So the only point to be tested is , which of course converges.
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