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

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:
$S_n=\frac{a(1-r^n)}{1-r}$
Given series,
$0.3+0.03+0.003+\cdots$
Since
$0.3=0.3$
$0.03=0.3(0.1)$
$0.003=0.3(0.1{)}^2$
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
$0.9=0.9$
$0.09=0.9(0.1)$
$0.009=0.9(0.1{)}^2$
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:
$S_n=\frac{a(1-r^n)}{1-r}$
Since $a=0.9,r=0.1$
Sum of the first term,
$S_1=\frac{0.9(1-(0.1{)}^1)}{1-(0.1)}=\frac{0.9(0.9)}{0.9}=0.9$
Sum of the first two terms,
$S_2=\frac{0.9(1-(0.1{)}^2)}{1-(0.1)}=\frac{0.9(0.99)}{0.9}=0.99$
Sum of the first three terms,
$S_3=\frac{0.9(1-(0.1{)}^3)}{1-(0.1)}=\frac{0.9(0.999)}{0.9}=0.999$
Sum of the first four terms,
$S_4=\frac{0.9(1-(0.1{)}^4)}{1-(0.1)}=\frac{0.9(0.9999)}{0.9}=0.9999$
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
$0.9+0.09+0.009+0.0009+...=1$

Answer is given below (on video)