How do you determine the convergence or divergence of #sum_(n=1)^oo (-1)^n/((2n-1)!)#?

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Answer 1 · 2017-07-03T19:23:23

One can use the Ratio Test
#lim_(ntooo) |a_(n+1)/a_n|#

Given: #a_n = (-1)^n/((2n-1)!)#

#a_(n+1) = (-1)^(n+1)/((2n+1-1)!)#

#a_(n+1)/a_n = ((-1)^(n+1)/((2n+1-1)!))/((-1)^n/((2n-1)!))#

#a_(n+1)/a_n = ((-1)(-1)^(n)/((2n+1-1)!))/((-1)^n/((2n-1)!))#

#a_(n+1)/a_n = ((-1)/((2n+1-1)!))/((1)/((2n-1)!))#

#a_(n+1)/a_n = ((-1)/((2n)!))/((1)/((2n-1)!))#

#a_(n+1)/a_n = ((-1)/((2n)(2n-1)!))/((1)/((2n-1)!))#

#a_(n+1)/a_n = (-1(2n-1)!)/((2n)(2n-1)!)#

#a_(n+1)/a_n = -1/(2n)#

#lim_(ntooo) |a_(n+1)/a_n| = lim_(ntooo) 1/(2n) = 0#

This absolutely converges.