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High School Math : Finding Sums of Infinite Series

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Example Questions

Example Question #2061 : High School Math

Find the value for $\sum_{i=0}^{\infty} (\frac{1}{2})^i$ $\displaystyle \sum_{i=0}^{\infty} (\frac{1}{2})^i$

Possible Answers:

$\displaystyle 1$

$\displaystyle 2$

1.25 $\displaystyle 1.25$

1.5 $\displaystyle 1.5$

$\displaystyle 0$

Correct answer:

$\displaystyle 2$

Explanation:

To best understand, let's write out the series. So

$\sum_{i=0}^{\infty} (\frac{1}{2})^i = 1+1/2+(1/2)^2 + (1/2)^3+...+(1/2)^\infty$ $\displaystyle \sum_{i=0}^{\infty} (\frac{1}{2})^i = 1+1/2+(1/2)^2 + (1/2)^3+...+(1/2)^\infty$

We can see this is an infinite geometric series with each successive term being multiplied by $\frac{1}{2}$ $\displaystyle \frac{1}{2}$.

A definition you may wish to remember is

$1+r+r^2+r^3+... +r^\infty= 1/(1-r)$ $\displaystyle 1+r+r^2+r^3+... +r^\infty= 1/(1-r)$ where $\displaystyle r$ stands for the common ratio between the numbers, which in this case is $\frac{1}{2}$ $\displaystyle \frac{1}{2}$ or $\displaystyle .5$. So we get

$\displaystyle 1/(1-r) = 1/(1-.5) = 1/.5 = 2$

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Example Question #31 : Sequences And Series

Evaluate:

$1 + \frac{1}{7} + \frac{1}{49} + \frac{1}{343} +...$ $\displaystyle 1 + \frac{1}{7} + \frac{1}{49} + \frac{1}{343} +...$

Possible Answers:

The series does not converge.

$\frac{4}{3}$ $\displaystyle \frac{4}{3}$

$\frac{6}{5}$ $\displaystyle \frac{6}{5}$

$\frac{9}{8}$ $\displaystyle \frac{9}{8}$

$\frac{7}{6}$ $\displaystyle \frac{7}{6}$

Correct answer:

$\frac{7}{6}$ $\displaystyle \frac{7}{6}$

Explanation:

This is a geometric series whose first term is $a_{0} = 1$ $\displaystyle a_{0} = 1$ and whose common ratio is $r = \frac{1}{7}$ $\displaystyle r = \frac{1}{7}$. The sum of this series is:

$\frac{a_{0}}{1-r} = \frac{1}{1- \frac{1}{7} } = \frac{1}{\frac{7}{7}- \frac{1}{7}}= \frac{1}{\frac{6}{7}} =\frac{7}{6}$ $\displaystyle \frac{a_{0}}{1-r} = \frac{1}{1- \frac{1}{7} } = \frac{1}{\frac{7}{7}- \frac{1}{7}}= \frac{1}{\frac{6}{7}} =\frac{7}{6}$

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Example Question #2 : Finding Sums Of Infinite Series

Evaluate:

$1 - \frac{1}{6} + \frac{1}{36} - \frac{1}{216} +...$ $\displaystyle 1 - \frac{1}{6} + \frac{1}{36} - \frac{1}{216} +...$

Possible Answers:

$\frac{7}{8}$ $\displaystyle \frac{7}{8}$

$\frac{2}{3}$ $\displaystyle \frac{2}{3}$

$\frac{6}{7}$ $\displaystyle \frac{6}{7}$

The series does not converge.

$\frac{4}{5}$ $\displaystyle \frac{4}{5}$

Correct answer:

$\frac{6}{7}$ $\displaystyle \frac{6}{7}$

Explanation:

This is a geometric series whose first term is $a_{0} = 1$ $\displaystyle a_{0} = 1$ and whose common ratio is $r = - \frac{1}{6}$ $\displaystyle r = - \frac{1}{6}$. The sum of this series is:

$\frac{a_{0}}{1-r} = \frac{1}{1- \left (- \frac{1}{6} \right )} = \frac{1}{\frac{6}{6}+ \frac{1}{6}}= \frac{1}{\frac{7}{6}} =\frac{6}{7}$ $\displaystyle \frac{a_{0}}{1-r} = \frac{1}{1- \left (- \frac{1}{6} \right )} = \frac{1}{\frac{6}{6}+ \frac{1}{6}}= \frac{1}{\frac{7}{6}} =\frac{6}{7}$

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High School Math : Finding Sums of Infinite Series

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #2061 : High School Math

Example Question #31 : Sequences And Series

Example Question #2 : Finding Sums Of Infinite Series

All High School Math Resources

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