Algebra II - Infinite Series

Evaluate the infinite series for

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:

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

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

$\frac{12}{7}$

$-\frac{1}{6}$

$\textup{The series will diverge.}$

$\frac{8}{5}$

$\frac{5}{3}$

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 $\frac{12}{7}$ .

Determine the sum, if possible: $4-\frac{2}{3}+\frac{1}{9}-...$

$\frac{24}{7}$

$\textup{The sum cannot be determined.}$

$\frac{71}{21}$

$\frac{17}{5}$

$\frac{64}{19}$

Explanation

Determine the common ratio of the infinite series by dividing the second term with the first term and the third term with the second term. The common ratios should be similar.

$\frac{-\frac{2}{3}}{4} = -\frac{2}{3} \div 4= -\frac{2}{3}\times\frac{1}{4} = -\frac{1}{6}$

$\frac{\frac{1}{9}}{-\frac{2}{3}} = \frac{1}{9} \div -\frac{2}{3} = \frac{1}{9} \times -\frac{3}{2} = -\frac{1}{6}$

Write the formula for infinite series, and substitute the terms.

$\textup{Sum} =\frac{\textup{first term}}{1-r} = \frac{4}{1-(-\frac{1}{6})} = \frac{4}{\frac{7}{6}}$

Simplify the complex fraction.

$\frac{4}{\frac{7}{6}} = 4\div \frac{7}{6} = 4 \times \frac{6}{7} = \frac{24}{7}$

The answer is: $\frac{24}{7}$

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

$\frac{27}{2}$

$\frac{29}{2}$

$\frac{40}{3}$

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: $\frac{27}{2}$

Find the sum of the infinite series

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.

Evaluate:

$\sum_{k = 1}^{\infty} \frac{1}{k^{2}+k}$

$\frac{1}{2}$

The series diverges.

Explanation

A trick to working this is to rewrite the expression

$\frac{1}{k^{2}+k}$

by, first, factoring the denominator:

$= \frac{1}{k (k+1)}$

Using the method of partial fractions, we can rewrite this as

$= \frac{A}{k } + \frac{B}{k+1}$

$= \frac{A (k+1)}{k (k+1) } + \frac{B k}{k (k+1)}$

$= \frac{A k+A}{k (k+1) } + \frac{B k}{k (k+1)}$

$= \frac{(A+B) k+A}{k (k+1) }$

Comparing numerators, we get

and

and the series can be restated as

$\sum_{k = 1}^{\infty} \left ( \frac{1}{k} - \frac{1}{k+1} \right )$

This can be rewritten by substituting the integers from 1 to 10, in turn, and adding:

$=\left ( 1 - \frac{1}{2}\right ) + \left ( \frac{1}{2} - \frac{1}{3}\right )+ \left ( \frac{1}{3} - \frac{1}{4}\right )+ ... .$

Regrouping, we see this is a telescoping series, in which all numbers after 1 cancel out:

$=1+\left ( - \frac{1}{2} + \frac{1}{2} \right ) +\left ( - \frac{1}{3} + \frac{1}{3} \right )+\left ( - \frac{1}{4} + \frac{1}{4} \right ) +...$

Evaluate:

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

$\frac{4}{3}$

$\frac{7}{4}$

$\frac{6}{7}$

The series diverges

Explanation

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

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

Evaluate:

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

$\frac{5}{8}$

$\frac{3}{8}$

$-\frac{5}{8}$

$-\frac{3}{8}$

The series diverges.

Explanation

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

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

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

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

$= \frac{5}{8}$

Evaluate:

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

The series diverges

$\frac{5}{9}$

$-\frac{5}{9}$

$\frac{4}{5}$

$-\frac{4}{5}$

Explanation

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.

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

$\frac{25}{4}$

$\frac{28}{3}$

$\frac{19}{3}$

$\textup{The series will diverge.}$

Explanation

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: $\frac{25}{4}$

Infinite Series

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