How do you use the integral test to determine if #Sigma lnn/n^3# from #[2,oo)# is convergent or divergent?

1 Answer
Andrea S.
Dec 12, 2016

#sum_(n=2)^(+oo) lnn/n^3# is convergent

Explanation:

We can start noting that for #n=1#, #a_n = ln n/n^3=0#, so we can write the series as:

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

We now that this series is convergent iff the integral:

#int_1^(+oo) (lnx )/x^3dx#

is convergent as well.

We can calculate the integral by parts:

#int_1^(+oo) (lnx )/x^3dx = -1/2int_1^(+oo) lnx* d(1/x^(2)) = [-1/2 ln x/x^2]_1^(+oo) +1/2 int_1^(+oo)1/x*1/x^2dx = 0 + 1/2 int_1^(+oo)(dx)/x^3 = [-1/(4x^2)]_1^(+oo)=1/4#

and therefore the series is convergent.

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