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Algebra II : Infinite Series

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Example Question #11 : Infinite Series

Find the sum if the series converges: $10-6+\frac{18}{5}-\frac{54}{25}+...$

Possible Answers:

$\frac{25}{4}$

$\textup{The series will diverge.}$

$\frac{28}{3}$

$\frac{19}{3}$

Correct answer:

$\frac{25}{4}$

Explanation:

Write the formula for finding the sum of an infinite geometric series.

$S=\frac{a_1}{1-r}$

The first term a_1 is ten.

Find the common ratio by dividing the second term with the first term, third term with the second term, or the fourth term with the third term, and so forth.

The common ratio should all be the same after dividing each term.

$r=\frac{-6}{10} = -\frac{3}{5}$

As long as is between negative one and one, we can use the formula to find the sum. Substitute the givens into the equation and solve.

$S=\frac{10}{1-(-\frac{3}{5})} = \frac{10}{\frac{8}{5}} = 10\times \frac{5}{8}= \frac{50}{8}= \frac{25}{4}$

The answer is: $\frac{25}{4}$

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Example Question #12 : Infinite Series

Determine the sum, if the series converges: $2+\frac{2}{9}+\frac{2}{81}+...$

Possible Answers:

$\frac{9}{4}$

$\frac{7}{2}$

$\frac{7}{4}$

$\frac{11}{4}$

$\frac{5}{2}$

Correct answer:

$\frac{9}{4}$

Explanation:

Write the formula for an infinite series.

$S=\frac{a_1}{1-r}$

The term a_1 represents the first term. The value of the common ratio must be between negative one and positive one, and can be determined by dividing the second term with the first term, third term with the second term, and so forth.

$\frac{2}{9}\div 2 = \frac{2}{9} \times \frac{1}{2} =\frac{1}{9}$

$\frac{2}{81}\div\frac{2}{9} = \frac{2}{81}\times\frac{9}{2} = \frac{9}{81} = \frac{1}{9}$

We can see that the common ratio is: $r=\frac{1}{9}$

Substitute the known values into the formula.

$S=\frac{2}{1-\frac{1}{9}} = \frac{2}{\frac{8}{9}} = 2\div \frac{8}{9} = 2\times \frac{9}{8}= \frac{9}{4}$

The sum will converge to: $\frac{9}{4}$

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Example Question #13 : Infinite Series

Determine the sum of: $9+3+1+\frac{1}{3}+\frac{1}{9}+...$

Possible Answers:

$\frac{29}{2}$

$\frac{40}{3}$

$\frac{27}{2}$

Correct answer:

$\frac{27}{2}$

Explanation:

Notice that this an infinite series.

$9+3+1+\frac{1}{3}+\frac{1}{9}+...$

Find the common ratio by dividing the second term with the first term, third term with the second term, and so forth. The common ratio should be same for each term divided.

$\frac{3}{9}=\frac{\frac{1}{3}}{1} = \frac{1}{3}$

The common ratio is: $r=\frac{1}{3}$

Write the formula for the sum of an infinite series.

$S=\frac{\textup{first term}}{1-r} = \frac{9}{1-\frac{1}{3}} = \frac{9}{\frac{2}{3}} = \frac{27}{2}$

The answer is: $\frac{27}{2}$

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Example Question #14 : Infinite Series

Determine the sum of the infinite series: $1+\frac{5}{6}+\frac{25}{36}+...$

Possible Answers:

$\frac{50}{3}$

$\textup{The series will diverge.}$

$\frac{75}{9}$

Correct answer:

Explanation:

To find the sum, write the formula for infinite series.

$S = \frac{\textup{first term}}{1-r}$

Determine the common ratio by dividing the second term with the first term, or third term with the second term. They should have similar common ratios.

Simplify the complex fractions to verify that both have similar common ratios.

$\frac{(\frac{5}{6})}{1} = \frac{5}{6}$

$\frac{\frac{25}{36}}{\frac{5}{6}} = \frac{25}{36}\div \frac{5}{6} = \frac{25}{36}\times \frac{6}{5} = \frac{5}{6}$

Substitute the known terms into the formula.

$S = \frac{1}{1-\frac{5}{6}} = \frac{1}{\frac{6}{6}-\frac{5}{6}} = \frac{1}{\frac{1}{6}}$

Simplify the complex fraction.

$\frac{1}{\frac{1}{6}} = 1 \div \frac{1}{6} = 1\times 6 = 6$

The sum will converge to .

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Example Question #15 : Infinite Series

Calculate the sum of the following series, if it exists:

$6+2+\frac{2}{3}+\frac{2}{9}+...$

Possible Answers:

Correct answer:

Explanation:

Each successive term in this series is determined by multiplying the previous term by $\frac{1}{3}$ ; hence, the series is geometric and its common ratio is $r=\frac{1}{3}$ .

Since $|r|=\frac{1}{3}<1$ , the sum of this series exists and we can calculate it via the following formula:

$S=\frac{a_1}{1-r}$ ,

where is the desired sum of the geometric series, is the common ratio of the geometric series, and a_1 is the first term appearing in the series.

Here, $r=\frac{1}{3}$ and a_1=6 . Hence, the sum of this geometric series is

$S=\frac{6}{1-\frac{1}{3}}=\frac{6}{\frac{2}{3}}=\frac{(6)(3)}{2}=9$ .

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Example Question #16 : Infinite Series

Determine the sum: $1-\frac{1}{11}+\frac{1}{121}-...$

Possible Answers:

$\frac{121}{12}$

$\frac{11}{10}$

$\frac{11} {12}$

$\frac{6}{5}$

$\frac{12}{11}$

Correct answer:

$\frac{11} {12}$

Explanation:

Write the formula for the sum of an infinite series.

$\textup{Sum} = \frac{\textup{first term}}{1-r}$

The is the common ratio between the terms in the series. This can be found by dividing the second term with the first term, the third term with the second, and so forth.

$\frac{-\frac{1}{11}}{1} =-\frac{1}{11}$

$\frac{\frac{1}{121}}{-\frac{1}{11}}=\frac{1}{121} \div -\frac{1}{11} = \frac{1}{121} \times -\frac{11}{1}=-\frac{1}{11}$

Substitute the known terms into the equation.

$\textup{Sum} = \frac{1}{1-(-\frac{1}{11})} = \frac{1}{1+\frac{1}{11}} =\frac{1}{\frac{11}{11}+\frac{1}{11}} =\frac{1}{\frac{12}{11}}$

Simplify the complex fraction.

$\frac{1}{\frac{12}{11}} = 1\div \frac{12}{11} = 1\times \frac{11} {12} = \frac{11} {12}$

The answer will converge to: $\frac{11} {12}$

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Example Question #17 : Infinite Series

Determine the sum, if possible: $2-\frac{1}{3}+\frac{1}{18}-...$

Possible Answers:

$\frac{12}{7}$

$\textup{The series will diverge.}$

$\frac{5}{3}$

$-\frac{1}{6}$

$\frac{8}{5}$

Correct answer:

$\frac{12}{7}$

Explanation:

Write the formula for the sum of an infinite series.

$\textup{Sum} =\frac{\textup{first term}}{1-r}$

The first term is two. To determine the common ratio, we will need to divide the second term by the first, third term with the second, and so forth. The common ratio should be same for each term.

$r=\frac{-\frac{1}{3}}{2} = -\frac{1}{3}\div 2 = -\frac{1}{3} \times \frac{1}{2} = -\frac{1}{6}$

$r=\frac{\frac{1}{18}}{-\frac{1}{3}} = \frac{1}{18}\div -\frac{1}{3} = \frac{1}{18}\times -\frac{3}{1} =-\frac{1}{6}$

The common ratios are verified to be the same. Substitute the into the formula. This value must be between negative one and one or the series will diverge!

$\textup{Sum} =\frac{\textup{first term}}{1-r} = \frac{2}{1-(-\frac{1}{6})} = \frac{2}{\frac{7}{6}}$

Simplify this complex fraction.

$2\div \frac{7}{6} =2\times \frac{6}{7} =\frac{12}{7}$

The series will converge to $\frac{12}{7}$ .

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Example Question #18 : Infinite Series

Determine the sum, if possible: $1+\frac{1}{8}+\frac{1}{64}+...$

Possible Answers:

$\textup{The series will diverge.}$

$\frac{8}{7}$

$\frac{4}{3}$

$\frac{8}{5}$

$\frac{12}{11}$

Correct answer:

$\frac{8}{7}$

Explanation:

Write the formula for the sum of an infinite series.

$\textup{Sum} =\frac{\textup{first term}}{1-r}$

Determine the common ratio, . Divide the second term by the first. This ratio should be the same if the third term was divided by the second term, and so forth.

$\frac{\frac{1}{8}}{1} = \frac{1}{8}$

The common ratio is $\frac{1}{8}$ .

Substitute the known terms into the equation.

$\textup{Sum} =\frac{1}{1-\frac{1}{8}} = \frac{1}{\frac{7}{8}}$

Solve the complex fraction.

$\frac{1}{\frac{7}{8}} = 1\div \frac{7}{8} = 1\times \frac{8}{7} = \frac{8}{7}$

The answer is: $\frac{8}{7}$

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Example Question #11 : Infinite Series

Find the sum, if possible: $5+\frac{3}{5}+\frac{9}{125}+...$

Possible Answers:

$\textup{The series will diverge.}$

$\frac{125}{3}$

$\frac{57}{10}$

$\frac{125}{22}$

$\frac{709}{125}$

Correct answer:

$\frac{125}{22}$

Explanation:

Write the formula for infinite series.

$\textup{Sum} =\frac{\textup{first term}}{1-r}$

Determine the common ratio by dividing the second term with the first term.

$\frac{3}{5}\div 5 = \frac{3}{5}\times \frac{1}{5} = \frac{3}{25}$

The ratio should also be similar if the third term is divided by the second term. Verify that the common ratios are similar.

$\frac{9}{125}\div \frac{3}{5} = \frac{9}{125}\times \frac{5}{3} = \frac{3}{25}$

Substitute the known terms into the equation.

$\textup{Sum} =\frac{\textup{first term}}{1-r} = \frac{5}{1-\frac{3}{25}} = \frac{5}{\frac{25}{25}-\frac{3}{25}}= \frac{5}{\frac{22}{25}}$

Simplify the complex fraction.

$5\div \frac{22}{25} = 5\times \frac{25}{22} = \frac{125}{22}$

The sum will converge to: $\frac{125}{22}$

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Example Question #20 : Infinite Series

Determine the sum, if possible: $3-\frac{1}{4}+\frac{1}{48}-...$

Possible Answers:

$\textup{The series will diverge.}$

$\frac{17}{6}$

$\frac{61}{23}$

$\frac{31}{11}$

$\frac{36}{13}$

Correct answer:

$\frac{36}{13}$

Explanation:

Find the common ratio of the infinite series. Divide the second term with the first, or the third with the second term. The common ratios should be similar to each other.

$-\frac{1}{4} \div 3 = -\frac{1}{4} \times \frac{1}{3} = -\frac{1}{12}$

$\frac{1}{48}\div -\frac{1}{4} =\frac{1}{48}\times- \frac{4}{1}=- \frac{1}{12}$

Write the formula for infinite series and substitute the numbers.

$\textup{Sum} =\frac{\textup{first term}}{1-r}=\frac{3}{1-(-\frac{1}{12})}$

Simplify the denominator.

$\frac{3}{\frac{12}{12}+\frac{1}{12}} = \frac{3}{\frac{13}{12}}$

$\frac{3}{\frac{13}{12}} =3\times \frac{12}{13} = \frac{36}{13}$

The sum converges to $\frac{36}{13}$ .

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Algebra II : Infinite Series

Study concepts, example questions & explanations for Algebra II

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