Answer :
To find the sum of the infinite series
[tex]\[ \frac{2}{3} + \frac{3}{3^2} + \frac{2}{3^3} + \frac{3}{3^4} + \frac{2}{3^5} + \frac{3}{3^6} + \ldots \][/tex]
we can observe that it alternates between terms of the form [tex]\(\frac{2}{3^n}\)[/tex] and [tex]\(\frac{3}{3^n}\)[/tex].
To simplify, we can split this into two separate geometric series:
1. Series 1: [tex]\(\frac{2}{3} + \frac{2}{3^3} + \frac{2}{3^5} + \ldots\)[/tex]
2. Series 2: [tex]\(\frac{3}{3^2} + \frac{3}{3^4} + \frac{3}{3^6} + \ldots\)[/tex]
### Analyzing Series 1:
Let's start with Series 1:
[tex]\[ \frac{2}{3} + \frac{2}{3^3} + \frac{2}{3^5} + \ldots \][/tex]
We can factor out the common factor of 2 from each term, giving:
[tex]\[ 2 \left(\frac{1}{3} + \frac{1}{3^3} + \frac{1}{3^5} + \ldots\right) \][/tex]
This is a geometric series where the first term [tex]\(a_1 = \frac{2}{3}\)[/tex] and the common ratio [tex]\(r_1 = \left(\frac{1}{3}\right)^2 = \frac{1}{9}\)[/tex]. The sum [tex]\(S_1\)[/tex] of an infinite geometric series with first term [tex]\(a\)[/tex] and common ratio [tex]\(r\)[/tex] (where [tex]\(|r| < 1\)[/tex]) is given by:
[tex]\[ S = \frac{a}{1-r} \][/tex]
Thus, for Series 1:
[tex]\[ a_1 = \frac{2}{3}, \quad r_1 = \frac{1}{9} \][/tex]
[tex]\[ S_1 = \frac{\frac{2}{3}}{1 - \frac{1}{9}} = \frac{\frac{2}{3}}{\frac{8}{9}} = \frac{2}{3} \cdot \frac{9}{8} = \frac{2 \cdot 9}{3 \cdot 8} = \frac{18}{24} = \frac{3}{4} = 0.75 \][/tex]
### Analyzing Series 2:
Now, let's consider Series 2:
[tex]\[ \frac{3}{3^2} + \frac{3}{3^4} + \frac{3}{3^6} + \ldots \][/tex]
Similarly, we can factor out the common factor of 3 from each term, yielding:
[tex]\[ 3 \left(\frac{1}{3^2} + \frac{1}{3^4} + \frac{1}{3^6} + \ldots\right) \][/tex]
This is another geometric series where the first term [tex]\(a_2 = \frac{3}{9} = \frac{1}{3}\)[/tex] and the common ratio [tex]\(r_2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9}\)[/tex]. Thus:
[tex]\[ a_2 = \frac{1}{3}, \quad r_2 = \frac{1}{9} \][/tex]
[tex]\[ S_2 = \frac{\frac{1}{3}}{1 - \frac{1}{9}} = \frac{\frac{1}{3}}{\frac{8}{9}} = \frac{1}{3} \cdot \frac{9}{8} = \frac{9}{24} = \frac{3}{8} = 0.375 \][/tex]
### Adding the Two Series:
Finally, to find the total sum of the original series, we add the sums of the two separate series:
[tex]\[ S_{\text{total}} = S_1 + S_2 = 0.75 + 0.375 = 1.125 \][/tex]
Thus, the sum of the infinite series [tex]\(\frac{2}{3} + \frac{3}{3^2} + \frac{2}{3^3} + \frac{3}{3^4} + \frac{2}{3^5} + \frac{3}{3^6} + \ldots\)[/tex] is
[tex]\[ \boxed{1.125} \][/tex]
[tex]\[ \frac{2}{3} + \frac{3}{3^2} + \frac{2}{3^3} + \frac{3}{3^4} + \frac{2}{3^5} + \frac{3}{3^6} + \ldots \][/tex]
we can observe that it alternates between terms of the form [tex]\(\frac{2}{3^n}\)[/tex] and [tex]\(\frac{3}{3^n}\)[/tex].
To simplify, we can split this into two separate geometric series:
1. Series 1: [tex]\(\frac{2}{3} + \frac{2}{3^3} + \frac{2}{3^5} + \ldots\)[/tex]
2. Series 2: [tex]\(\frac{3}{3^2} + \frac{3}{3^4} + \frac{3}{3^6} + \ldots\)[/tex]
### Analyzing Series 1:
Let's start with Series 1:
[tex]\[ \frac{2}{3} + \frac{2}{3^3} + \frac{2}{3^5} + \ldots \][/tex]
We can factor out the common factor of 2 from each term, giving:
[tex]\[ 2 \left(\frac{1}{3} + \frac{1}{3^3} + \frac{1}{3^5} + \ldots\right) \][/tex]
This is a geometric series where the first term [tex]\(a_1 = \frac{2}{3}\)[/tex] and the common ratio [tex]\(r_1 = \left(\frac{1}{3}\right)^2 = \frac{1}{9}\)[/tex]. The sum [tex]\(S_1\)[/tex] of an infinite geometric series with first term [tex]\(a\)[/tex] and common ratio [tex]\(r\)[/tex] (where [tex]\(|r| < 1\)[/tex]) is given by:
[tex]\[ S = \frac{a}{1-r} \][/tex]
Thus, for Series 1:
[tex]\[ a_1 = \frac{2}{3}, \quad r_1 = \frac{1}{9} \][/tex]
[tex]\[ S_1 = \frac{\frac{2}{3}}{1 - \frac{1}{9}} = \frac{\frac{2}{3}}{\frac{8}{9}} = \frac{2}{3} \cdot \frac{9}{8} = \frac{2 \cdot 9}{3 \cdot 8} = \frac{18}{24} = \frac{3}{4} = 0.75 \][/tex]
### Analyzing Series 2:
Now, let's consider Series 2:
[tex]\[ \frac{3}{3^2} + \frac{3}{3^4} + \frac{3}{3^6} + \ldots \][/tex]
Similarly, we can factor out the common factor of 3 from each term, yielding:
[tex]\[ 3 \left(\frac{1}{3^2} + \frac{1}{3^4} + \frac{1}{3^6} + \ldots\right) \][/tex]
This is another geometric series where the first term [tex]\(a_2 = \frac{3}{9} = \frac{1}{3}\)[/tex] and the common ratio [tex]\(r_2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9}\)[/tex]. Thus:
[tex]\[ a_2 = \frac{1}{3}, \quad r_2 = \frac{1}{9} \][/tex]
[tex]\[ S_2 = \frac{\frac{1}{3}}{1 - \frac{1}{9}} = \frac{\frac{1}{3}}{\frac{8}{9}} = \frac{1}{3} \cdot \frac{9}{8} = \frac{9}{24} = \frac{3}{8} = 0.375 \][/tex]
### Adding the Two Series:
Finally, to find the total sum of the original series, we add the sums of the two separate series:
[tex]\[ S_{\text{total}} = S_1 + S_2 = 0.75 + 0.375 = 1.125 \][/tex]
Thus, the sum of the infinite series [tex]\(\frac{2}{3} + \frac{3}{3^2} + \frac{2}{3^3} + \frac{3}{3^4} + \frac{2}{3^5} + \frac{3}{3^6} + \ldots\)[/tex] is
[tex]\[ \boxed{1.125} \][/tex]
