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Sure, let's tackle each problem step-by-step.
### Problem 1: Convergence of the Series
Given the series:
[tex]\[ 1 - \frac{1}{\sqrt{3}} + \frac{1}{3} - \frac{1}{3\sqrt{3}} + \frac{1}{9} - \frac{1}{9\sqrt{3}} + \cdots \][/tex]
To determine whether this series converges or diverges, and if it converges, find the sum, we need to identify whether it represents a geometric series.
#### Step 1: Identify the Pattern
Observe the terms of the series and how they change. We can group the series in pairs of positive and negative terms as follows:
[tex]\[ \left(1 - \frac{1}{\sqrt{3}}\right) + \left(\frac{1}{3} - \frac{1}{3\sqrt{3}}\right) + \left(\frac{1}{9} - \frac{1}{9\sqrt{3}}\right) + \cdots \][/tex]
#### Step 2: Recognize the Series as Geometric
Each pair of terms can be expressed as:
[tex]\[ \left(\frac{1}{3^n} - \frac{1}{3^n \cdot \sqrt{3}}\right) \][/tex]
Factor out [tex]\( \frac{1}{3^n} \)[/tex] from each term:
[tex]\[ \left(\frac{1}{3^n} \times 1 - \frac{1}{3^n} \times \frac{1}{\sqrt{3}}\right) = \frac{1}{3^n} \left(1 - \frac{1}{\sqrt{3}}\right) \][/tex]
#### Step 3: Simplify the Series
Let:
[tex]\[ a = 1 - \frac{1}{\sqrt{3}} \][/tex]
Then our series becomes:
[tex]\[ a \left(1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots\right) \][/tex]
#### Step 4: Sum the Geometric Series
This is a geometric series with the first term [tex]\( a \)[/tex] and common ratio [tex]\( r = \frac{1}{3} \)[/tex]. The sum [tex]\( S \)[/tex] of an infinite geometric series [tex]\( a + ar + ar^2 + ar^3 + \cdots \)[/tex] can be found using the formula:
[tex]\[ S = \frac{a}{1 - r} \][/tex]
Here, [tex]\( a = 1 - \frac{1}{\sqrt{3}} \)[/tex] and [tex]\( r = \frac{1}{3} \)[/tex]. Plugging these in:
[tex]\[ S = \frac{1 - \frac{1}{\sqrt{3}}}{1 - \frac{1}{3}} = \frac{1 - \frac{1}{\sqrt{3}}}{\frac{2}{3}} = \frac{3}{2} \left(1 - \frac{1}{\sqrt{3}}\right) \][/tex]
Thus, the series converges and the sum is:
[tex]\[ \boxed{\frac{3}{2} \left(1 - \frac{1}{\sqrt{3}}\right)} \][/tex]
### Problem 2: Fractional Expansion of Repeating Decimal
Given the repeating decimal:
[tex]\[ 0.424242 \overline{42} \][/tex]
Let's convert this repeating decimal into a fraction.
#### Step 1: Express the Repeating Decimal as a Geometric Series
Notice that:
[tex]\[ 0.42424242\ldots = 0.42 + 0.0042 + 0.000042 + \cdots \][/tex]
We can write this as:
[tex]\[ 0.42 (1 + 0.01 + 0.0001 + \cdots) \][/tex]
Notice this forms a geometric series where each term is [tex]\( 0.01 \)[/tex] times the previous term.
#### Step 2: Write it as a Geometric Series
Let:
[tex]\[ a = 0.42 = \frac{42}{100} \][/tex]
[tex]\[ r = 0.01 = \frac{1}{100} \][/tex]
Thus:
[tex]\[ 0.424242\ldots = \frac{42}{100} \left(1 + \left(\frac{1}{100}\right) + \left(\frac{1}{100}\right)^2 + \cdots\right) \][/tex]
#### Step 3: Sum the Geometric Series
The sum [tex]\( S \)[/tex] of an infinite geometric series [tex]\( a + ar + ar^2 + ar^3 + \cdots \)[/tex] is:
[tex]\[ S = \frac{a}{1 - r} \][/tex]
Here, [tex]\( a = \frac{42}{100} \)[/tex] and [tex]\( r = \frac{1}{100} \)[/tex]:
[tex]\[ S = \frac{\frac{42}{100}}{1 - \frac{1}{100}} = \frac{\frac{42}{100}}{\frac{99}{100}} = \frac{42}{99} \][/tex]
#### Step 4: Simplify the Fraction
Simplify [tex]\( \frac{42}{99} \)[/tex] by dividing the numerator and the denominator by their greatest common divisor, which is 3:
[tex]\[ \frac{42 \div 3}{99 \div 3} = \frac{14}{33} \][/tex]
Thus, the fractional expansion of the repeating decimal [tex]\( 0.424242 \overline{42} \)[/tex] is:
[tex]\[ \boxed{\frac{14}{33}} \][/tex]
### Problem 1: Convergence of the Series
Given the series:
[tex]\[ 1 - \frac{1}{\sqrt{3}} + \frac{1}{3} - \frac{1}{3\sqrt{3}} + \frac{1}{9} - \frac{1}{9\sqrt{3}} + \cdots \][/tex]
To determine whether this series converges or diverges, and if it converges, find the sum, we need to identify whether it represents a geometric series.
#### Step 1: Identify the Pattern
Observe the terms of the series and how they change. We can group the series in pairs of positive and negative terms as follows:
[tex]\[ \left(1 - \frac{1}{\sqrt{3}}\right) + \left(\frac{1}{3} - \frac{1}{3\sqrt{3}}\right) + \left(\frac{1}{9} - \frac{1}{9\sqrt{3}}\right) + \cdots \][/tex]
#### Step 2: Recognize the Series as Geometric
Each pair of terms can be expressed as:
[tex]\[ \left(\frac{1}{3^n} - \frac{1}{3^n \cdot \sqrt{3}}\right) \][/tex]
Factor out [tex]\( \frac{1}{3^n} \)[/tex] from each term:
[tex]\[ \left(\frac{1}{3^n} \times 1 - \frac{1}{3^n} \times \frac{1}{\sqrt{3}}\right) = \frac{1}{3^n} \left(1 - \frac{1}{\sqrt{3}}\right) \][/tex]
#### Step 3: Simplify the Series
Let:
[tex]\[ a = 1 - \frac{1}{\sqrt{3}} \][/tex]
Then our series becomes:
[tex]\[ a \left(1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots\right) \][/tex]
#### Step 4: Sum the Geometric Series
This is a geometric series with the first term [tex]\( a \)[/tex] and common ratio [tex]\( r = \frac{1}{3} \)[/tex]. The sum [tex]\( S \)[/tex] of an infinite geometric series [tex]\( a + ar + ar^2 + ar^3 + \cdots \)[/tex] can be found using the formula:
[tex]\[ S = \frac{a}{1 - r} \][/tex]
Here, [tex]\( a = 1 - \frac{1}{\sqrt{3}} \)[/tex] and [tex]\( r = \frac{1}{3} \)[/tex]. Plugging these in:
[tex]\[ S = \frac{1 - \frac{1}{\sqrt{3}}}{1 - \frac{1}{3}} = \frac{1 - \frac{1}{\sqrt{3}}}{\frac{2}{3}} = \frac{3}{2} \left(1 - \frac{1}{\sqrt{3}}\right) \][/tex]
Thus, the series converges and the sum is:
[tex]\[ \boxed{\frac{3}{2} \left(1 - \frac{1}{\sqrt{3}}\right)} \][/tex]
### Problem 2: Fractional Expansion of Repeating Decimal
Given the repeating decimal:
[tex]\[ 0.424242 \overline{42} \][/tex]
Let's convert this repeating decimal into a fraction.
#### Step 1: Express the Repeating Decimal as a Geometric Series
Notice that:
[tex]\[ 0.42424242\ldots = 0.42 + 0.0042 + 0.000042 + \cdots \][/tex]
We can write this as:
[tex]\[ 0.42 (1 + 0.01 + 0.0001 + \cdots) \][/tex]
Notice this forms a geometric series where each term is [tex]\( 0.01 \)[/tex] times the previous term.
#### Step 2: Write it as a Geometric Series
Let:
[tex]\[ a = 0.42 = \frac{42}{100} \][/tex]
[tex]\[ r = 0.01 = \frac{1}{100} \][/tex]
Thus:
[tex]\[ 0.424242\ldots = \frac{42}{100} \left(1 + \left(\frac{1}{100}\right) + \left(\frac{1}{100}\right)^2 + \cdots\right) \][/tex]
#### Step 3: Sum the Geometric Series
The sum [tex]\( S \)[/tex] of an infinite geometric series [tex]\( a + ar + ar^2 + ar^3 + \cdots \)[/tex] is:
[tex]\[ S = \frac{a}{1 - r} \][/tex]
Here, [tex]\( a = \frac{42}{100} \)[/tex] and [tex]\( r = \frac{1}{100} \)[/tex]:
[tex]\[ S = \frac{\frac{42}{100}}{1 - \frac{1}{100}} = \frac{\frac{42}{100}}{\frac{99}{100}} = \frac{42}{99} \][/tex]
#### Step 4: Simplify the Fraction
Simplify [tex]\( \frac{42}{99} \)[/tex] by dividing the numerator and the denominator by their greatest common divisor, which is 3:
[tex]\[ \frac{42 \div 3}{99 \div 3} = \frac{14}{33} \][/tex]
Thus, the fractional expansion of the repeating decimal [tex]\( 0.424242 \overline{42} \)[/tex] is:
[tex]\[ \boxed{\frac{14}{33}} \][/tex]
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