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 of0.3+0.03+0.003+...?

Question
Series
asked 2021-02-03

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 \(0.9 + 0.09 + 0.009 +\) ⋅ ⋅ ⋅?

Answers (1)

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
\(0.9+0.09+0.009+0.0009+...=1\)

0

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