EXAMPLE 5 Write the number $3.9 \overline{21}=3.9212121 \ldots$ as a ratio of integers.

SOLUTION
$$
3.9212121 \ldots=3.9+\frac{21}{10^{3}}+\frac{21}{10^{5}}+\frac{21}{10^{7}}+\cdots
$$

After the first term we have a geometric series with $a=\frac{21}{10^{3}}$ and $r=1 / 100 \quad$. Therefore
$$
\begin{aligned}
3.9 \overline{21} & =3.9+\frac{\frac{21}{10^{3}}}{1-1 / 100} \
& =3.9+\frac{\frac{21}{\frac{99}{100}}}{100} \
& =\frac{39}{10}+\frac{21}{990} \
& =\square
\end{aligned}
$$
Calculate the sum of the series $\sum_{n=1}^{\infty} a_{n}$ whose partial sums are given.
$$
s_{n}=9-7(0.7)^{n}
$$

Question

EXAMPLE 5 Write the number $3.9 \overline{21}=3.9212121 \ldots$ as a ratio of integers.

SOLUTION
$$
3.9212121 \ldots=3.9+\frac{21}{10^{3}}+\frac{21}{10^{5}}+\frac{21}{10^{7}}+\cdots
$$

After the first term we have a geometric series with $a=\frac{21}{10^{3}}$ and $r=1 / 100 \quad$. Therefore
$$
\begin{aligned}
3.9 \overline{21} & =3.9+\frac{\frac{21}{10^{3}}}{1-1 / 100} \
& =3.9+\frac{\frac{21}{\frac{99}{100}}}{100} \
& =\frac{39}{10}+\frac{21}{990} \
& =\square
\end{aligned}
$$
Calculate the sum of the series $\sum_{n=1}^{\infty} a_{n}$ whose partial sums are given.
$$
s_{n}=9-7(0.7)^{n}
$$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To convert the repeating decimal $3.9\overline{21}$ to a ratio of integers, we start by expressing it as a sum of a whole number and a geometric series: $3.9212121\ldots = 3.9 + \frac{21}{10^3} + \frac{21}{10^5} + \frac{21}{10^7} + \cdots$ ‖ ‖ step 2 ⋮ Identify the first term $a$ and the common ratio $r$ of the geometric series. Here, $a = \frac{21}{10^3}$ and $r = \frac{1}{100}$. ‖ ‖ step 3 ⋮ Use the formula for the sum of an infinite geometric series: $\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}$. Therefore, $3.9\overline{21} = 3.9 + \frac{\frac{21}{10^3}}{1 - \frac{1}{100}}$. ‖ ‖ step 4 ⋮ Simplify the expression: $3.9 + \frac{\frac{21}{10^3}}{\frac{99}{100}} = 3.9 + \frac{21}{990}$. ‖ ‖ step 5 ⋮ Convert $3.9$ to a fraction: $\frac{39}{10}$. ‖ ‖ step 6 ⋮ Combine the fractions: $\frac{39}{10} + \frac{21}{990} = \frac{39 \cdot 99 + 21}{990} = \frac{3861 + 21}{990} = \frac{3882}{990} = \frac{1941}{495}$. ‖ ∻Answer∻ ⚹ $\frac{1941}{495}$ ⚹ ∻Key Concept∻ ⚹Converting repeating decimals to fractions⚹ ∻Explanation∻ ⚹The key concept is to express the repeating decimal as a sum of a whole number and a geometric series, then use the formula for the sum of an infinite geometric series to find the ratio of integers.⚹ ∻Solution by Steps∻ ‖ step 1 ⋮ To find the sum of the series $\sum_{n=1}^{\infty} a_n$ with partial sums $s_n = 9 - 7(0.7)^n$, we first note that the series converges if the partial sums approach a finite limit as $n$ approaches infinity. ‖ ‖ step 2 ⋮ Calculate the limit of the partial sums: $\lim_{n o \infty} s_n = \lim_{n o \infty} (9 - 7(0.7)^n)$. ‖ ‖ step 3 ⋮ Since $0.7^n$ approaches $0$ as $n$ approaches infinity, we have $\lim_{n o \infty} (9 - 7(0.7)^n) = 9 - 7 \cdot 0 = 9$. ‖ ∻Answer∻ ⚹ $9$ ⚹ ∻Key Concept∻ ⚹Convergence of series with given partial sums⚹ ∻Explanation∻ ⚹The key concept is to determine the limit of the partial sums as $n$ approaches infinity. If the limit exists and is finite, the series converges to that limit.⚹◊What is the formula for the sum of an infinite geometric series?⍭ Generate me a similar question◊

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