Questions: Using the limit comparison test with a geometric comparison series of sum from n=1 to infinity of 2^n / 7^n, determine if the series below converges or diverges. sum from n=1 to infinity of (2^n + 9) / 7^n

Solution Steps

To determine if the series \(\sum_{n=1}^{\infty} \frac{2^{n}+9}{7^{n}}\) converges or diverges, we can use the limit comparison test with the geometric series \(\sum_{n=1}^{\infty} \frac{2^{n}}{7^{n}}\). The geometric series \(\sum_{n=1}^{\infty} \frac{2^{n}}{7^{n}}\) is known to converge because its common ratio \(\frac{2}{7}\) is less than 1. We will compute the limit of the ratio of the terms of the given series and the comparison series as \(n\) approaches infinity. If the limit is a positive finite number, then both series either converge or diverge together.

We are given the series

\[ \sum_{n=1}^{\infty} \frac{2^{n}+9}{7^{n}} \]

and we will compare it with the geometric series

\[ \sum_{n=1}^{\infty} \frac{2^{n}}{7^{n}}. \]

The comparison series can be expressed as

\[ \sum_{n=1}^{\infty} \left(\frac{2}{7}\right)^{n}. \]

Since the common ratio \(\frac{2}{7} < 1\), this series converges.

We compute the limit of the ratio of the terms of the two series:

\[ \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{2^{n}+9}{7^{n}}}{\frac{2^{n}}{7^{n}}} = \lim_{n \to \infty} \frac{2^{n}+9}{2^{n}} = \lim_{n \to \infty} \left(1 + \frac{9}{2^{n}}\right). \]

As \(n\) approaches infinity, \(\frac{9}{2^{n}} \to 0\), thus:

\[ \lim_{n \to \infty} \left(1 + \frac{9}{2^{n}}\right) = 1. \]

Since the limit is a positive finite number (\(1\)), and the comparison series converges, we conclude that the original series

\[ \sum_{n=1}^{\infty} \frac{2^{n}+9}{7^{n}} \]

also converges.

Final Answer