How do you test the series #Sigma (n+3)/(n(n+1)(n-2))# from n is #[3,oo)# for convergence?

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Answer 1 · 2017-02-09T14:44:36

#sum_(n=3)^oo (n+3)/(n(n+1)(n-2))#

is convergent.

Given:

#sum_(n=3)^oo a_n = sum_(n=3)^oo (n+3)/(n(n+1)(n-2))#

we can use the direct comparison test, identifying a convergent series #b_n# such that:

#a_n < b_n# for #n > N#

We can start from:

#a_n = (n+3)/(n(n+1)(n-2))#

if we increase the numerator and decrease the denominator, the resulting quotient will be bigger:

#a_n = (n+3)/(n(n+1)(n-2)) = (n-2)/(n(n+1)(n-2)) + 5/(n(n+1)(n-2)) = 1/(n(n+1)) + 5/(n(n+1)(n-2))#

#a_n < 1/n^2 +5/n^2 = 6/n^2#

As #sum_(n=3)^oo 1/n^2# is convergent based on the p-series test, then also:

#sum_(n=3)^oo (n+3)/(n(n+1)(n-2))#

is convergent.