How do you test the series #Sigma (100^n(n!)^3)/((3n)!)# from #n=[1,oo)# by the ratio test?

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Answer 1 · 2017-01-25T15:52:36

The series:

#sum_(n=1)^oo((100^n(n!)^3)/((3n)!))#

is divergent.

The ratio test states that given the series:

#sum_(n=1)^oo a_n#

and the limit:

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

the series is absolutely convergent if #L < 1# and divergent if # L > 1#.

Let's calculate the ratio for our series:

# abs (a_(n+1)/a_n) = abs ( ( (100^(n+1)((n+1)!)^3)/((3(n+1))!)) /((100^n(n!)^3)/((3n)!))) #

# abs (a_(n+1)/a_n) = ( 100^(n+1)/100^n) (((n+1)!)^3/ (n!)^3) ( ((3n)!)/((3n+3)!))#

# abs (a_(n+1)/a_n) = 100 (((n+1)!)/(n!))^3 ( ((3n)!) / ( (3n+3)(3n+2)(3n+1) (3n)!))#

# abs (a_(n+1)/a_n) = (100 (n+1)^3)/( (3n+3)(3n+2)(3n+1))#

So:

#lim_(n->oo) abs (a_(n+1)/a_n) = 100/27 > 1#

and the series is divergent.