How do you use the direct comparison test to determine if #Sigma 3^n/(4^n+5)# from #[0,oo)# is convergent or divergent?

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Answer 1 · 2018-02-01T08:23:47

See explanation.

#4^n + 5 > 4^n# which means that #1/(4^n+5) < 1/4^n#.

Since #3^n >0#, #3^n/(4^n+5) < 3^n/4^n# for all #n>=0#.

Given this, we know #sum3^n/(4^n+5) < sum3^n/4^n#.

#sum(3/4)^n# is a convergent geometric series.

Since #sum3^n/(4^n+5)# is positive termed and #sum3^n/(4^n+5) < sum3^n/4^n#, #sum3^n/(4^n+5)# must converge by direct comparison.