Finding Sums of Infinite Series

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

CORRECT

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

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

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

A definition you may wish to remember is

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

Evaluate:

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

$\frac{7}{6}$

CORRECT

The series does not converge.

$\frac{6}{5}$

$\frac{4}{3}$

$\frac{9}{8}$

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

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

Evaluate:

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

$\frac{6}{7}$

CORRECT

The series does not converge.

$\frac{7}{8}$

$\frac{4}{5}$

$\frac{2}{3}$

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

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