The following series are geometric series or a sum of two geometric series. Determine whether each series converges or not. For the series which converge, enter the sum of the series. For the series which diverges enter "DIV" (without quotes). (a) $$ \sum_{n=1}^{\infty} \frac{10^{n}}{9^{n}}= $$ (b) $$ \sum_{n=2}^{\infty} \frac{1}{3^{n}}= $$ (c) $$ \sum_{n=0}^{\infty} \frac{3^{n}}{4^{2 n+1}}= $$ (d) $$ \sum_{n=5}^{\infty} \frac{9^{n}}{1^{n}}= $$ (e) $$ \sum_{n=1}^{\infty} \frac{5^{n}}{5^{n+4}}= $$ (f) $$ \sum_{n=1}^{\infty} \frac{9^{n}+3^{n}}{10^{n}}= $$

Question

UpStudy AI · Accepted Answer

To determine whether each series converges or diverges, we will analyze each series individually. A geometric series converges if the absolute value of the common ratio $$ r $$ is less than 1, and the sum can be calculated using the formula $$ S = \frac{a}{1 - r} $$, where $$ a $$ is the first term. Let's analyze each series step by step: ### (a) $$ \sum_{n=1}^{\infty} \frac{10^{n}}{9^{n}} $$ This can be rewritten as: $$ \sum_{n=1}^{\infty} \left( \frac{10}{9} \right)^{n} $$ Here, the common ratio $$ r = \frac{10}{9} $$. Since $$ |r| > 1 $$, the series diverges. **Answer:** DIV ### (b) $$ \sum_{n=2}^{\infty} \frac{1}{3^{n}} $$ This can be rewritten as: $$ \sum_{n=2}^{\infty} \left( \frac{1}{3} \right)^{n} $$ Here, the common ratio $$ r = \frac{1}{3} $$. Since $$ |r| < 1 $$, the series converges. The first term $$ a = \frac{1}{3^2} = \frac{1}{9} $$. The sum is: $$ S = \frac{a}{1 - r} = \frac{\frac{1}{9}}{1 - \frac{1}{3}} = \frac{\frac{1}{9}}{\frac{2}{3}} = \frac{1}{9} \cdot \frac{3}{2} = \frac{1}{6} $$ **Answer:** $$ \frac{1}{6} $$ ### (c) $$ \sum_{n=0}^{\infty} \frac{3^{n}}{4^{2 n+1}} $$ This can be rewritten as: $$ \sum_{n=0}^{\infty} \frac{3^{n}}{4^{2n} \cdot 4} = \frac{1}{4} \sum_{n=0}^{\infty} \left( \frac{3}{16} \right)^{n} $$ Here, the common ratio $$ r = \frac{3}{16} $$. Since $$ |r| < 1 $$, the series converges. The first term $$ a = 1 $$. The sum is: $$ S = \frac{a}{1 - r} = \frac{1}{4} \cdot \frac{1}{1 - \frac{3}{16}} = \frac{1}{4} \cdot \frac{16}{13} = \frac{4}{13} $$ **Answer:** $$ \frac{4}{13} $$ ### (d) $$ \sum_{n=5}^{\infty} \frac{9^{n}}{1^{n}} $$ This can be rewritten as: $$ \sum_{n=5}^{\infty} 9^{n} $$ Here, the common ratio $$ r = 9 $$. Since $$ |r| > 1 $$, the series diverges. **Answer:** DIV ### (e) $$ \sum_{n=1}^{\infty} \frac{5^{n}}{5^{n+4}} $$ This can be simplified to: $$ \sum_{n=1}^{\infty} \frac{1}{5^{4}} = \frac{1}{625} \sum_{n=1}^{\infty} 1 $$ This series diverges because it is a constant term summed infinitely. **Answer:** DIV ### (f) $$ \sum_{n=1}^{\infty} \frac{9^{n}+3^{n}}{10^{n}} $$ This can be separated into two series: $$ \sum_{n=1}^{\infty} \left( \frac{9^{n}}{10^{n}} + \frac{3^{n}}{10^{n}} \right) = \sum_{n=1}^{\infty} \left( \frac{9}{10} \right)^{n} + \sum_{n=1}^{\infty} \left( \frac{3}{10} \right)^{n} $$ Both series have common ratios $$ r_1 = \frac{9}{10} $$ and $$ r_2 = \frac{3}{10} $$, both of which are less than 1, so both series converge. The sums are: 1. For $$ r_1 = \frac{9}{10} $$: $$ S_1 = \frac{\frac{9}{10}}{1 - \frac{9}{10}} = \frac{\frac{9}{10}}{\frac{1}{10}} = 9 $$ 2. For $$ r_2 = \frac{3}{10} $$: $$ S_2 = \frac{\frac{3}{10}}{1 - \frac{3}{10}} = \frac{\frac{3}{10}}{\frac{7}{10}} = \frac{3}{7} $$ The total sum is: $$ S = S_1 + S_2 = 9 + \frac{3}{7} = \frac{63}{7} + \frac{3}{7} = \frac{66}{7} $$ **Answer:** $$ \frac{66}{7} $$ ### Summary of Answers: (a) DIV (b) $$ \frac{1}{6} $$ (c) $$ \frac{4}{13} $$ (d) DIV (e) DIV (f) $$ \frac{66}{7} $$ (a) DIV (b) $$ \frac{1}{6} $$ (c) $$ \frac{4}{13} $$ (d) DIV (e) DIV (f) $$ \frac{66}{7} $$

Upstudy AI Solution

Answer

Solution

Beyond the Answer

Related Questions

Latest Calculus Questions