Pick out the conveyent geometcii seies and Find its sum :

Calculus: Early Transcendentals
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Chapter1: Functions And Models
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**Selecting and Summing Convergent Geometric Series**

The goal is to identify the convergent geometric series from the following options and then determine its sum:

1. \( \sum_{n=1}^{\infty} \frac{1}{n} \)

2. \( \sum_{n=1}^{\infty} 2(1.01)^n \)

3. \( \sum_{n=1}^{\infty} 2(0.99)^{n-1} \)

**Explanation:**

- The first series is the harmonic series, which is known to diverge.

- The second series has a common ratio of 1.01, which is greater than 1, making it a divergent geometric series.

- The third series has a common ratio of 0.99, which is less than 1, indicating that it is a convergent geometric series.

To calculate the sum of the convergent geometric series \( \sum_{n=1}^{\infty} 2(0.99)^{n-1} \), use the formula for the sum of an infinite geometric series:

\[ S = \frac{a}{1 - r} \]

where \( a = 2 \) is the first term and \( r = 0.99 \) is the common ratio.

Thus, the sum is:

\[ S = \frac{2}{1 - 0.99} = \frac{2}{0.01} = 200 \]
Transcribed Image Text:**Selecting and Summing Convergent Geometric Series** The goal is to identify the convergent geometric series from the following options and then determine its sum: 1. \( \sum_{n=1}^{\infty} \frac{1}{n} \) 2. \( \sum_{n=1}^{\infty} 2(1.01)^n \) 3. \( \sum_{n=1}^{\infty} 2(0.99)^{n-1} \) **Explanation:** - The first series is the harmonic series, which is known to diverge. - The second series has a common ratio of 1.01, which is greater than 1, making it a divergent geometric series. - The third series has a common ratio of 0.99, which is less than 1, indicating that it is a convergent geometric series. To calculate the sum of the convergent geometric series \( \sum_{n=1}^{\infty} 2(0.99)^{n-1} \), use the formula for the sum of an infinite geometric series: \[ S = \frac{a}{1 - r} \] where \( a = 2 \) is the first term and \( r = 0.99 \) is the common ratio. Thus, the sum is: \[ S = \frac{2}{1 - 0.99} = \frac{2}{0.01} = 200 \]
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