To examine the sum of all the terms of an infinite geometric sequence, we need to consider $S_n = \dfrac{u_1(1-r^n)}{1-r}$ when $n$ gets very large.
If $\left|r\right|>1$, the series is said to be divergent and the sum infinitely large.
For instance, when $r=2$ and $u_1=1$;
$S_\infty=1+2+4+8+\cdots$ is infinitely large.
If $\left|r\right|<1$, or $-1 \lt r \lt 1$, then as $n$ becomes very large, $r^n$ approaches $0$.
For instance, when $r=\dfrac{1}{2}$ and $u_1=1$;
$S_\infty=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\cdots = 2$.
$$S_\infty=\dfrac{u_1}{1-r}$$
We call this the limiting sum of the series.
This result can be used to find the value of recurring decimals.
Let’s take a look at $0.\overline{7}$ to see how to convert it to a fraction.
\( \begin{align} \displaystyle
0.\overline{7} &= 0.7 + 0.07 + 0.007 + 0.0007 + \cdots \\
&= 0.7 + 0.7(0.1) + 0.7(0.1)^2 + 0.7(0.1)^3 + \cdots \\
&= \dfrac{0.7}{1-0.1} \\
&= \dfrac{0.7}{0.9} \\
&= \dfrac{7}{9} \\
\end{align} \)
Example 1
Write $0.\overline{12}$ as a rational number.
\( \begin{align} \displaystyle
0.\overline{12} &= 0.12 + 0.0012 + 0.000012 + \cdots \\
&= 0.12 + 0.12(0.1)^2 + 0.12(0.1)^4 + \cdots \\
&= \dfrac{0.12}{1-0.1^2} \\
&= \dfrac{0.12}{0.99} \\
&= \dfrac{4}{33}
\end{align} \)
Example 2
Write $0.1 \overline{2}$ as a rational number.
\( \begin{align} \displaystyle
0.1 \overline{2} &= 0.1222 + \cdots \\
&= 0.1 + 0.02 + 0.002 + 0.0002 + \cdots \\
&= 0.1 + 0.02 + 0.02(0.1) + 0.02(0.1)^2 + \cdots \\
&= 0.1 + \dfrac{0.02}{1-0.1} \\
&= 0.1 + \dfrac{0.02}{0.9} \\
&= \dfrac{11}{90}
\end{align} \)
Example 3
Write $1.2 \overline{345}$ as a rational number.
\( \begin{align} \displaystyle 1.2 \overline{345} &= 1.2345345345 + \cdots \\ &= 1.2 + 0.0345 + 0.0000345 + 0.0000000345 + \cdots \\ &= 1.2 + 0.0345 + 0.0345(0.001)+ 0.0345(0.001)^2 + \cdots \\ &= 1.2 + \dfrac{0.0345}{1-0.001} \\ &= 1 \dfrac{731}{3330} \end{align} \)
Example 4
\( \displaystyle S=\sum^{\infty}_{n=1} \frac{3}{5^n} \)
(a) Determine if the series \( \mathcal{S} \) is convergent or divergent. State reason(s). Note the starting value of \( n \) is \( 1 \).
\( \displaystyle \begin{align} \text{the first term} &= \frac{3}{5} \\ \text{common ratio} &= \frac{1}{5} \\ \therefore \mathcal{S} \text{ is convergent as } &-1 \lt \text{common ratio} \lt 1 \end{align} \)
(b) If convergent, find the value of the series.
\( \displaystyle \begin{align} \mathcal{S}_{\infty} &= \frac{\text{first term}}{1-\text{common ratio}} \\ &= \frac{\displaystyle \frac{3}{5}}{1-\displaystyle \frac{1}{5}} \\ &= \frac{\displaystyle \frac{3}{5}}{\displaystyle \frac{4}{5}} \\ &= \frac{3}{4} \end{align} \)
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