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

Question

Emilia Aguilar · Accepted Answer

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.

Isaiah Allen · Answer

undefined To determine if the series $ \sum_{n=0}^{\infty} \frac{3^n}{4^n+5} $ converges or diverges, we can use the Direct Comparison Test.

First, observe that for $ n \geq 0 $:

$$ \frac{3^n}{4^n + 5} < \frac{3^n}{4^n} = \left(\frac{3}{4}\right)^n $$

Since the series $ \sum_{n=0}^{\infty} \left(\frac{3}{4}\right)^n $ is a geometric series with $ |r| = \frac{3}{4} < 1 $, it converges.

Now, by the Direct Comparison Test, since $ 0 \leq \frac{3^n}{4^n + 5} < \left(\frac{3}{4}\right)^n $ and $ \sum_{n=0}^{\infty} \left(\frac{3}{4}\right)^n $ converges, then $ \sum_{n=0}^{\infty} \frac{3^n}{4^n + 5} $ also converges.

Therefore, the series $ \sum_{n=0}^{\infty} \frac{3^n}{4^n + 5} $ is convergent.