# Reasoning as in the given problem, what is the value of0.3+0.03+0.003+...?

Reasoning as in the given problem, what is the value of
$0.3+0.03+0.003+...?$
Working with series Consider the infinite series
$0.9+0.09+0.009+0.0009+...,$
where each term of the sum is $\frac{1}{10}$ of the previous term.
a. Find the sum of the first one, two, three, and four terms of the series.
b. What value would you assign to the infinite series $0.9+0.09+0.009+$ ⋅ ⋅ ⋅?

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Latisha Oneil

Sum of infinite geometric series:
Any series in the form of $a,ar,a{r}^{2},...$ is called infinite geometric series and sum of infinite geometry series is,
${S}_{\mathrm{\infty }}=\frac{a}{1-r}$
Where a is the first term and r is a common ratio.
And Sum of geometric series:
${S}_{n}=\frac{a\left(1-{r}^{n}\right)}{1-r}$
Given series,
$0.3+0.03+0.003+\cdots$
Since
$0.3=0.3$
$0.03=0.3\left(0.1\right)$
$0.003=0.3\left(0.1{\right)}^{2}$
So given series is a geometric series with first term $a=0.3$ and common ratio is 0.1.
So sum of infinite geometry series is,
${S}_{\mathrm{\infty }}=\frac{0.3}{1-0.1}$
${S}_{\mathrm{\infty }}=\frac{0.3}{0.9}$
${S}_{\mathrm{\infty }}=\frac{1}{3}$
${S}_{\mathrm{\infty }}=0.333...$
Hence
$0.3+0.03+0.003+...=0.333...$
Given series,
$0.9+0.09+0.009+0.0009+...$
Since
$0.9=0.9$
$0.09=0.9\left(0.1\right)$
$0.009=0.9\left(0.1{\right)}^{2}$
So given series is a geometric series with first term $a=0.9$ and common ratio is 0.1.
a) Evaluate the sum of the first one, two, three, and four terms of the series.
Since sum of geometric series:
${S}_{n}=\frac{a\left(1-{r}^{n}\right)}{1-r}$
Since $a=0.9,r=0.1$
Sum of the first term,
${S}_{1}=\frac{0.9\left(1-\left(0.1{\right)}^{1}\right)}{1-\left(0.1\right)}=\frac{0.9\left(0.9\right)}{0.9}=0.9$
Sum of the first two terms,
${S}_{2}=\frac{0.9\left(1-\left(0.1{\right)}^{2}\right)}{1-\left(0.1\right)}=\frac{0.9\left(0.99\right)}{0.9}=0.99$
Sum of the first three terms,
${S}_{3}=\frac{0.9\left(1-\left(0.1{\right)}^{3}\right)}{1-\left(0.1\right)}=\frac{0.9\left(0.999\right)}{0.9}=0.999$
Sum of the first four terms,
${S}_{4}=\frac{0.9\left(1-\left(0.1{\right)}^{4}\right)}{1-\left(0.1\right)}=\frac{0.9\left(0.9999\right)}{0.9}=0.9999$
Hence
Sum of the first term $=0.9$
Sum of the first two terms $=0.99$
Sum of the first three terms $=0.999$
Sum of the first four terms $=0.9999$
b) So sum of infinite geometry series is,
${S}_{\mathrm{\infty }}=\frac{0.9}{1-0.1}$
${S}_{\mathrm{\infty }}=\frac{0.9}{0.9}$
${S}_{\mathrm{\infty }}=1$
Hence
$0.9+0.09+0.009+0.0009+...=1$

Jeffrey Jordon