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Infinite Geometric Series


Predict $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\;...\;=$

Even though the series has an infinite number of terms, the sum is finite.

Exploring Infinite Series

If the sum of a series is infinite, we say it is d _ _ _ _ _ e n t.

If the sum of a series is finite, we say it is c _ _ _ _ _ _ e n t.
Arithmetic Series

For each of the following arithmetic series, write if it is “divergent” or “convergent”.

$[u_1=1, d=2]\quad1+3+5+7+9+\ldots=$

$[u_1=1, d=1]\quad1+2+3+4+5+\ldots=$

$[u_1=1, d=\frac{1}{2}]\quad1+1\frac{1}{2}+2+2\frac{1}{2}+3+\ldots=$

$[u_1=1, d=0]\quad1+1+1+1+1+\ldots=$

$[u_1=1, d=-\frac{1}{2}]\quad1+\frac{1}{2}+0-\frac{1}{2}-1-1\frac{1}{2}+\ldots=$

$[u_1=1, d=-1]\quad1+0-1-2-3+\ldots=$

$[u_1=1, d=-2]\quad1-1-3-5-7+\ldots=$

$[u_1=100, d=-1]\quad100+99+\ldots+1+0-1-2-3+\ldots=$

All infinite arithmetic series are
*Write "divergent" or "convergent"
Geometric Series

For each of the following geometric series, write if it is “divergent” or “convergent”.

$[u_1=1, r=2]\quad1+2+4+8+16+\ldots=$

$[u_1=1, r=1]\quad1+1+1+1+1+\ldots=$

$[u_1=1, r=\frac{1}{2}]\quad1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots=$

$[u_1=1, r=0]\quad1+0+0+0+0+\ldots=$

$[u_1=1, r=-\frac{1}{2}]\quad1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\ldots=$

$[u_1=1, r=-1]\quad1-1+1-1+1-\ldots=$

$[u_1=1, r=-2]\quad1-2+4-8+16-\ldots=$
Infinite geometric series are convergent if $\lt r \lt$

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

$2+\frac{4}{3}+\frac{8}{9}+\frac{16}{27}+\frac{32}{81}+\ldots=$

$\displaystyle{\sum_{n=1}^\infty \frac{3}{4^n}}=$

Exercises

(Core 5I on P.124) #1, 4c, 5b, 6, 8, 9, 10, 12