Question

consider the following series.
\\(\sum_{n = 1}^{\infty} 6^n + 17^{-n}\\)
determine whether the geometric series is convergent or divergent. justify your answer.

• converges; the series is a constant multiple of a geometric series.
• converges; the limit of the terms, \\(a_n\\), is 0 as \\(n\\) goes to infinity.
• diverges; the limit of the terms, \\(a_n\\), is not 0 as \\(n\\) goes to infinity.
• diverges; the series is a constant multiple of the harmonic series.

if it is convergent, find the sum. (if the quantity diverges, enter diverges.)

Explanation:

Step1: Rewrite the series

We can rewrite the series \(\sum_{n = 1}^{\infty}6^{n+1}7^{-n}\) as \(\sum_{n = 1}^{\infty}6\times6^{n}7^{-n}=6\sum_{n = 1}^{\infty}(\frac{6}{7})^{n}\).

Step2: Recall the formula for the sum of a geometric series

The sum of a geometric series \(\sum_{n = 1}^{\infty}ar^{n - 1}=\frac{a}{1 - r}\) when \(|r|\lt1\). Our series \(6\sum_{n = 1}^{\infty}(\frac{6}{7})^{n}\) can be rewritten in the form of the standard geometric - series formula. Let \(m=n - 1\), then \(n=m + 1\). When \(n = 1\), \(m = 0\). So \(6\sum_{n = 1}^{\infty}(\frac{6}{7})^{n}=6\times\frac{6}{7}\sum_{m = 0}^{\infty}(\frac{6}{7})^{m}\).
Here \(a=\frac{6}{7}\) and \(r=\frac{6}{7}\). Since \(|r|=\frac{6}{7}\lt1\), the sum of \(\sum_{m = 0}^{\infty}(\frac{6}{7})^{m}=\frac{1}{1-\frac{6}{7}}\) (using the formula \(\sum_{n = 0}^{\infty}ar^{n}=\frac{a}{1 - r}\) with \(a = 1\)).

Step3: Calculate the sum

First, \(\sum_{m = 0}^{\infty}(\frac{6}{7})^{m}=\frac{1}{1-\frac{6}{7}}=\frac{1}{\frac{1}{7}} = 7\). Then \(6\times\frac{6}{7}\sum_{m = 0}^{\infty}(\frac{6}{7})^{m}=6\times\frac{6}{7}\times7\).

Answer:

\(36\)