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Algebra II : Infinite Series

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Example Questions

Example Question #1 : Infinite Series

Which of the following infinite series has a finite sum?

Possible Answers:

$\sum_{n=0}^{\infty } \left ( \frac{1}{2} \right )^{n}$

$\sum_{n=0}^{\infty } 3*5^{n}$

$\sum_{n=0}^{\infty } \left ( \frac{7}{3} \right )^{n}$

$\sum_{n=0}^{\infty } \left ( \frac{3}{2} \right )^{n}$

$\sum_{n=0}^{\infty } 1.15^{n}$

Correct answer:

$\sum_{n=0}^{\infty } \left ( \frac{1}{2} \right )^{n}$

Explanation:

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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Example Question #2 : Infinite Series

Evaluate:

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

Possible Answers:

The series diverges.

$-\frac{3}{8}$

$\frac{5}{8}$

$\frac{3}{8}$

$-\frac{5}{8}$

Correct answer:

$\frac{5}{8}$

Explanation:

The sum of an infinite series $\sum_{i = 0}^{\infty}a^{i}$ , where |a| < 1 , 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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Example Question #3 : Infinite Series

Evaluate:

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

Possible Answers:

$-\frac{4}{5}$

$\frac{5}{9}$

The series diverges

$-\frac{5}{9}$

$\frac{4}{5}$

Correct answer:

The series diverges

Explanation:

An infinite series $\sum_{i = 1}^{\infty}a^{i}$ converges to a sum if and only if |a|<1 . 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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Example Question #4 : Infinite Series

Evaluate:

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

Possible Answers:

$\frac{6}{7}$

$\frac{7}{4}$

The series diverges

$\frac{4}{3}$

Correct answer:

$\frac{4}{3}$

Explanation:

The sum of an infinite series $\sum_{i = 1}^{\infty}a^{i}$ , where |a| < 1 , 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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Example Question #1 : Infinite Series

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

Possible Answers:

$\frac{10}{3}$

$\frac{5}{2}$

$\frac{8}{3}$

Correct answer:

Explanation:

Write the formula for infinite geometric series.

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

The value of a_1 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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Example Question #6 : Infinite Series

If a_1= 5 and $r=\frac{1}{4}$ , what will be the sum of the infinite series?

Possible Answers:

$\frac{20}{3}$

$\frac{15}{4}$

$\text{Diverges}$

$\frac{19}{2}$

Correct answer:

$\frac{20}{3}$

Explanation:

Write the infinite series formula.

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

Substitute the values of a_1 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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Example Question #7 : Infinite Series

Evaluate the infinite series for 10 - 9 + 8.1 - 7.29 + ...

Possible Answers:

5.263

0.19

100

Correct answer:

5.263

Explanation:

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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Example Question #101 : Mathematical Relationships And Basic Graphs

Find the sum of the infinite series 10 + 11 + 12.1 + 13.31 + ...

Possible Answers:

100

Cannot be determined - the sum is infinite

Correct answer:

Cannot be determined - the sum is infinite

Explanation:

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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Example Question #9 : Infinite Series

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

Possible Answers:

$\frac{1}{2}$

$\frac{1}{4}$

$\frac{2}{5}$

$\textup{The series will diverge.}$

Correct answer:

$\frac{1}{4}$

Explanation:

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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Example Question #10 : Infinite Series

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

Possible Answers:

$\frac{2}{3}$

$\frac{8}{3}$

$\frac{3}{8}$

$\textup{The series does not converge.}$

$\frac{1}{8}$

Correct answer:

$\frac{8}{3}$

Explanation:

Write the formula for the sum of an infinite series.

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

The value a_1 is the first term, and is the common ratio.

a_1=4

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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Algebra II : Infinite Series

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #1 : Infinite Series

Example Question #2 : Infinite Series

Example Question #3 : Infinite Series

Example Question #4 : Infinite Series

Example Question #1 : Infinite Series

Example Question #6 : Infinite Series

Example Question #7 : Infinite Series

Example Question #101 : Mathematical Relationships And Basic Graphs

Example Question #9 : Infinite Series

Example Question #10 : Infinite Series

All Algebra II Resources

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