How do you use the limit comparison test to determine if #Sigma (5n-3)/(n^2-2n+5)# from #[1,oo)# is convergent or divergent?

Question

6127 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-30T13:06:15

#sum_(n=1)^oo (5n-3)/(n^2-2n+5)# is divergent

#(5n-3)/(n^2-2n+5+7n+5/4) < (5n-3)/(n^2-2n+5)# but

#(5n-3)/(n^2-2n+5+7n+5/4) = (5n-3)/(n+5/2)^2=5/(n+5/2)-(31/2)/(n+5/2)^2#

We know that #sum_(n=1)^oo1/n^2# is convergent and #1/(n+5/2)^2 < 1/n^2# so

#sum_(n=1)^oo(31/2)/(n+5/2)^2# is convergent. Now examining

#sum_(n=1)^oo5/(n+5/2) = sum_(n=3)^oo(5/(n+1/2)) gt 5sum_(n=4)^oo1/n#

and we know that #sum_(k=4)^oo 1/n# is divergent.

Concluding

#sum_(n=1)^oo (5n-3)/(n^2-2n+5)# is divergent