How do you test for convergence for #sum (3n+7)/(2n^2-n)# for n is 1 to infinity?

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Answer 1 · 2018-04-28T15:05:55

The series diverge

The series is

#sum_(n=1)^oo(3n+7)/(2n^2-n)=sum_(n=1)^oo(3n)/(2n^2-n) + sum_(n=1)^oo7/(2n^2-n)#

#sum_(n=1)^oo(3n)/(2n^2-n)=3sum_(n=1)^oo(1)/(2n-1)=3/2sum_(n=1)^oo1/(n)#

This series diverge.

#sum_(n=1)^oo7/(2n^2-n)=7sum_(n=1)^oo1/(2n^2-n)#

This series converge as

#int_1^oo(dn)/(2n^2-1)=[ln(|2n-1|)-ln(n)]_0^oo=ln(2)#

Conclusion :

#" diverge + converge " = " diverge"#