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Algebra II : Infinite Series

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Example Question #1 :Infinite Series

Which of the following infinite series has a finite sum?

Possible Answers:

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

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

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

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

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

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 ofin each term of the series) must be betweenand. 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:

$-\frac{5}{8}$

$\frac{3}{8}$

$\frac{5}{8}$

The series diverges.

$-\frac{3}{8}$

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:

$\frac{4}{5}$

$-\frac{4}{5}$

The series diverges

$\frac{5}{9}$

$-\frac{5}{9}$

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:

$\frac{7}{4}$

The series diverges

$\frac{6}{7}$

$\frac{4}{3}$

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 #5 :Infinite Series

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

Possible Answers:

$\frac{5}{2}$

$\frac{8}{3}$

$\frac{10}{3}$

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:

$\frac{19}{2}$

$\text{Diverges}$

$\frac{20}{3}$

$\frac{15}{4}$

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:

100

0.19

5.263

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 #8 :Infinite Series

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 $\left | r \right | < 1$ 其中r是常见的比率。我们可以找到通讯器n 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:

$\frac{1}{4}$

$\frac{1}{2}$

$\frac{2}{5}$

$\textup{The series will diverge.}$

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:

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

$\frac{8}{3}$

$\frac{1}{8}$

$\frac{3}{8}$

$\frac{2}{3}$

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, andis 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.

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

查nge 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 #5 :Infinite Series

Example Question #6 :Infinite Series

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