Problem 44 - Essential Calculus

Question

Use the Direct Comparison Test to determine the convergence or divergence of the series. $$ \sum_{n=0}^{\infty} \frac{3^{n}}{4^{n}+5} $$

Step-by-Step Solution

Answer

The series \(\sum_{n=0}^{\infty} \frac{3^{n}}{4^{n}+5}\) is CONVERGENT.

Step 1: Understanding the series

We are given the series \(\sum_{n=0}^{\infty} \frac{3^{n}}{4^{n}+5}\). We need to decide which series to compare it with. For this, it is beneficial to take note of the biggest term in the denominator, which is \(4^{n}\). Therefore, we will compare our series with \(\sum_{n=0}^{\infty} \frac{3^{n}}{4^{n}}\).

Step 2: Recognizing the comparison series

The comparison series \(\sum_{n=0}^{\infty} \frac{3^{n}}{4^{n}}\) is a geometric series where the ratio \(r\), which is \(\frac{3}{4}\), is less than 1. This means by the Geometric Series Test, this comparison series is a convergent series.

Step 3: Applying the Direct Comparison Test

The Direct Comparison test states that if 0 ≤ aₙ ≤ bₙ for all n sufficiently large and \(\sum_{n=1}^{\infty} bₙ\) is convergent, then \(\sum_{n=1}^{\infty} aₙ\) is also convergent. In this situation, aₙ would be \(\frac{3^{n}}{4^{n}+5}\), and bₙ would be \(\frac{3^{n}}{4^{n}}\), and since 0 ≤ \(\frac{3^{n}}{4^{n}+5}\) ≤ \(\frac{3^{n}}{4^{n}}\) for all n, and we have established that the series for bₙ is convergent, this means that the given series \(\sum_{n=0}^{\infty} \frac{3^{n}}{4^{n}+5}\) is also convergent by the Direct Comparison Test.