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

Series
ANSWERED 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 $$0.9 + 0.09 + 0.009 +$$ ⋅ ⋅ ⋅? 2021-02-04

Sum of infinite geometric series:
Any series in the form of $$a,ar,ar^2,...$$ is called infinite geometric series and sum of infinite geometry series is,
$$S_{\infty}=\frac{a}{1-r}$$
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,
$$S_{\infty}=\frac{0.3}{1-0.1}$$
$$S_{\infty}=\frac{0.3}{0.9}$$
$$S_{\infty}=\frac13$$
$$S_{\infty}=0.333...$$
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,
$$S_{\infty}=\frac{0.9}{1-0.1}$$
$$S_{\infty}=\frac{0.9}{0.9}$$
$$S_{\infty}=1$$
Hence
$$0.9+0.09+0.009+0.0009+...=1$$