How do you test the series #Sigma 5^n/(3^n+4^n)# from n is #[0,oo)# for convergence?

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Answer 1 · 2017-01-17T22:22:01

The series is divergent.

#5^n/(3^n+4^n)=1/((3/5)^2+(4/5)^n)# but

#1/(2(4/5)^2) < 1/((3/5)^2+(4/5)^n) < 1/(2(3/5)^2)# or

#1/2(5/4)^n < 1/((3/5)^2+(4/5)^n) < 1/2(5/3)^n# but

#5/4 > 1# and #5/3 > 1# so

#sum_(n=1)^oo(5/4)^n=oo# and #sum_(n=1)^oo(5/3)^n=oo#

so

#sum_(n=0)^oo5^n/(3^n+4^n)=oo# being divergent