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

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