Infinite Series - Algebra II

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Question

Which of the following infinite series has a finite sum?

Answer

For an infinite series to have a finite sum, the exponential term (the term being raised to the power of in each term of the series) must be between and . Otherwise, each term is larger than the previous term, causing the overall sum to grow without bounds towards infinity.

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Question

Evaluate:

$\sum_{i = 1}^{\infty} \left (\frac{4}{7} \right )^{i}$

Answer

The sum of an infinite series $\sum_{i = 1}^{\infty}a^{i}$ , where , can be calculated as follows:

$\sum_{i = 1}^{\infty} a^{i} = \frac{a }{1- a}$

Setting $a = \frac{4}{7}$ :

$\sum_{i = 1}^{\infty} \left (\frac{4}{7} \right )^{i} = \frac{\frac{4}{7}}{1-\frac{4}{7}}= \frac{\frac{4}{7}}{ \frac{3}{7} } = \frac{4}{7} \div \frac{3}{7} = \frac{4}{7} \cdot \frac{7} {3} = \frac{4}{3}$

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Question

Evaluate:

$\sum_{i = 0}^{\infty} \left (- \frac{5}{4} \right )^{i}$

Answer

An infinite series $\sum_{i = 1}^{\infty}a^{i}$ converges to a sum if and only if . However, in the series $\sum_{i = 1}^{\infty} \left (- \frac{5}{4} \right )^{i}$ , this is not the case, as $\left | - \frac{5}{4} \right |= \frac{5}{4} > 1$ . This series diverges.

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Question

Evaluate:

$\sum_{i = 0}^{\infty} \left (- \frac{3}{5} \right )^{i}$

Answer

The sum of an infinite series $\sum_{i = 0}^{\infty}a^{i}$ , where , can be calculated as follows:

$\sum_{i = 0}^{\infty} a^{i} = \frac{1 }{1- a}$

Setting $a = - \frac{3}{5}$ :

$\sum_{i = 0}^{\infty} \left (- \frac{3}{5} \right )^{i} = \frac{1 }{1- \left (- \frac{3}{5} \right )}$

$= \frac{1 }{1+ \frac{3}{5} }$

$= \frac{1 }{ \frac{8}{5} }$

$= \frac{5}{8}$

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Question

Evaluate: $\sum_{n=1}^{\infty }\left(\frac{3}{4}\right)^n$

Answer

Write the formula for infinite geometric series.

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

The value of is the first term of the series, which is $\frac{3}{4}$ .

The value of the common ratio, , is also $\frac{3}{4}$ .

The ratio is $\frac{3}{4}$ because if we were to write out the first few terms in the series we would see,

$\frac{3}{4}^1+\frac{3}{4}^2+ \frac{3}{4}^3+...$

each term is three fourths more than the previous term therefore, giving us the ratio.

Substitute the values into the equation and evaluate.

$S=\frac{\frac{3}{4}}{1-\frac{3}{4}} = \frac{\frac{3}{4}}{\frac{1}{4}} =\frac{3}{4}\cdot4 = 3$

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Question

If and $r=\frac{1}{4}$ , what will be the sum of the infinite series?

Answer

Write the infinite series formula.

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

Substitute the values of and .

$S=\frac{a_1}{1-r}= \frac{5}{1-\frac{1}{4}} = \frac{5}{\frac{3}{4}} = 5\cdot\frac{4}{3}=\frac{20}{3}$

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Question

Evaluate the infinite series for

Answer

The first term of this sequence is 10. To find the common ratio r, we can just divide the second term by the first: $-9 \div 10 = - 0.9$ . So "r" is -0.9. We can find the infinite sum using the formula $\frac{a}{1-r}$ where a is the first term and r is the common ratio:

$\frac{10}{1-(-0.9)} = \frac{10}{1+0.9 } = \frac{10}{1.9} \approx 5.263$

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Question

Find the sum of the infinite series

Answer

An infinite sum is only calculable if $\left | r \right | < 1$ where r is the common ratio. We can find the common ratio easily by dividing the second term by the first: $11 \div 10 = 1.1$ . This is greater than 1, so we can't find the infinite sum - it is infinite.

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Question

What is the sum? $\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+...$

Answer

Write the formula to find the sum of an infinite geometric series.

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

The first term is: $a_1 = \frac{1}{8}$ .

The common ratio is: $r=\frac{1}{2}$

Substitute the values into the formula.

$S=\frac{\frac{1}{8}}{1-\frac{1}{2}} = \frac{\frac{1}{8}}{\frac{1}{2}}$

Rewrite the complex fraction.

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

The sum will converge to $\frac{1}{4}$ .

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Question

What is the sum? $4-2+1-\frac{1}{2}+\frac{1}{4}-...$

Answer

Write the formula for the sum of an infinite series.

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

The value is the first term, and is the common ratio.

Divide the second term with the first term, third term and the second, and so forth, and we will get a common ratio of:

$r=-\frac{1}{2}$

Substitute the values into the formula.

$S=\frac{a_1}{1-r}= \frac{4}{1-(-\frac{1}{2})} = \frac{4}{\frac{3}{2}}$

Rewrite the complex fraction using a division sign.

$\frac{4}{\frac{3}{2}} = 4 \div \frac{3}{2}$

Change the division sign to a multiplication and take the reciprocal of the second term.

$4 \times \frac{2}{3} = \frac{8}{3}$

The series will converge to $\frac{8}{3}$ .

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Question

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

Answer

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

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

The first term 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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Question

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

Answer

Write the formula for an infinite series.

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

The term 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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Question

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

Answer

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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Question

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

Answer

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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Question

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

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

Answer

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 is the first term appearing in the series.

Here, $r=\frac{1}{3}$ and . 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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Question

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

Answer

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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Question

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

Answer

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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Question

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

Answer

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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Question

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

Answer

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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Question

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

Answer

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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