How do you apply the ratio test to determine if #Sigma (1*3*5* * * (2n-1))/(n!)^2# from #n=[1,oo)# is convergent to divergent?

1 Answer
Andrea S.
Jan 17, 2017

The series:

#sum_(n=1)^oo (prod_1^n 2k-1)/(n!)^2#

is convergent.

Explanation:

Write the series as:

#sum_(n=1)^oo (prod_1^n 2k-1)/(n!)^2#

The ratio test states that a sufficient condition for #sum_(n=1)^oo a_n# to converge is that:

#L = lim_(n->oo) abs(a_(n+1)/a_n) < 1#

We then evaluate the ratio for the series at hand:

#abs(a_(n+1)/a_n) = frac ((prod_1^(n+1) (2k-1))/((n+1)!^2)) ((prod_1^n (2k-1))/(n!)^2) = ( (prod_1^(n+1) (2k-1)) / (prod_1^n (2k-1)) ) ((n!^2)/((n+1)!^2)) = (2(n+1)-1)/((n+1)^2) = (2n+1)/((n+1)^2)#

so that:

#lim_(n->oo) abs(a_(n+1)/a_n) = 0#

which proves the series is convergent.

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