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(1.3) Infinite Geometric Series


Predict $\quad \displaystyle \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\cdots=$

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

If the sum of a series is finite, we say it is

Arithmetic Series

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

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

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

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

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

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

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

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

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

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”.

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

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

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

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

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

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

$[u_1=1, r=-2]\quad1-2+4-8+16-\cdots=$
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 $\textcolor{red}{r^n}$ term: $$S_\infty=\frac{u_1\left(\textcolor{red}{r^n}-1\right)}{r-1} \; or \; \frac{u_1\left(1-\textcolor{red}{r^n}\right)}{1-r}$$ 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 $\displaystyle S_\infty=$
$\displaystyle \frac{u_1}{1-r}$ where $\lvert r\rvert \lt 1$

Find the value of $\displaystyle \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\cdots$

Find the value of $\displaystyle \frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}-\frac{1}{32}+\cdots$
*Give your answer as a fraction

Find the value of $\displaystyle 2+\frac{4}{3}+\frac{8}{9}+\frac{16}{27}+\frac{32}{81}+\cdots$

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

Write the value of $\displaystyle 0.12+0.0012+0.000012+0.00000012+\cdots$ as a fraction.

The sum of the first three terms of a convergent infinite geometric series is $37$. The sum to infinity of the series is $64$. Find the general term in the form $u_n=\displaystyle u_1 \left(r\right)^{n-1}$.
$u_n=$

The sum of an infinite geometric series is $S_\infty =32$ and $S_3=28$.
Find the general term in the form $u_n=\displaystyle u_1 \left(r\right)^{n-1}$.
$u_n=$

For what values of $x$ does the geometric series $3-24x^3+192x^6-1536x^9+\cdots$ converge.
$\lvert x \rvert<$

A geometric series converges to $8$ and $u_2=-\frac{5}{2}$. Find the common ratio.
$r=$

The geometric series $a+ar+ar^2+\cdots$ has a sum of $7$, and the terms involving odd powers of $r$ have a sum of $3$. What is $a+r$?
$a+r=$