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

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Answer 1 · 2018-03-21T03:19:54

It's convergent. You may use the integral test to prove this.

The ratio test isn't the correct test. I would use the integral test.

We need to evaluate the integral

#=int_1^oo 1/n^3 dn#

#=lim_(t-> oo) int _1^t 1/n^3 dn#

#=lim_(t-> oo) [-1/2n^-2]_1^t#

#=lim_(t->oo) -1/2t^-2 - (-1/2(1)^-2)#

#=0 + 1/2#

#= 1/2#

Since #int_1^oo 1/n^3 dn# is convergent, than the series is also convergent.

Hopefully this helps!