Infinite geometric series are convergent if
$\lt r \lt$
a)
b)
Finding the Sum of an Infinite Geometric Series
Recall from the previous section that the formula for the sum of a geometric series is:
$S_n=$
For an infinite series, $n\rightarrow\infty$. Let’s focus on the $\color{red}{r^n}$ term:
$$S_\infty=\frac{u_1\left(\color{red}{r^n}-1\right)}{r-1} \; or \; \frac{u_1\left(1-\color{red}{r^n}\right)}{1-r}$$
If $r=\frac{1}{2}$, then
*Give your answer as a decimal$\quad r^2=$
*Give your answer as a decimal$\quad r^3=$
*Give your answer as a decimal$\quad r^4=$
*Give your answer as a decimal$\quad r^5=$
*Give your answer as a decimal$\quad r^6=$
As $n\rightarrow \infty$, what number does $r^n$ approach?
For an infinite geometric series with $-1 \lt r \lt 1$, as $n\rightarrow \infty, r^n\rightarrow$
Therefore, the formula becomes $S_\infty=$
Which can be simplified to $\displaystyle{S_\infty=\frac{u_1}{1-r}}$ where $\lvert r\rvert \lt 1$
Using the formula, find $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\ldots=$
$\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}-\frac{1}{32}+\ldots=$
*Give your answer as a fraction
