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Determine the sums of the following geometric series when they are convergent. $$1+\frac{1}{6}+\frac{1}{6^{2}}+\frac{1}{6^{3}}+\frac{1}{6^{4}} \cdots$$

Short Answer

Expert verified
The sum of the series is \( \frac{6}{5} \).

Step by step solution

01

Identify the series type

Recognize that the given series is a geometric series: \[1 + \frac{1}{6} + \frac{1}{6^2} + \frac{1}{6^3} + \frac{1}{6^4} + \cdots\]
02

Identify the common ratio

Find the common ratio (r) of the series. The common ratio is found by dividing any term by the previous term. Here,\[r = \frac{\frac{1}{6}}{1} = \frac{1}{6}.\]
03

Check convergence

A geometric series converges if the absolute value of the common ratio is less than 1. Check if \(|r| < 1\).Since \(|\frac{1}{6}| = \frac{1}{6} < 1\), the series is convergent.
04

Find the sum of the series

Use the formula for the sum of a convergent geometric series, \( S = \frac{a}{1 - r} \). Here, \( a = 1 \) and \( r = \frac{1}{6} \). Substitute these values into the formula:\[ S = \frac{1}{1 - \frac{1}{6}} = \frac{1}{\frac{5}{6}} = \frac{6}{5} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Convergent Series
A series is a sum of terms that follow a specific pattern. In mathematics, a series can either converge or diverge. When we say a series is convergent, it means that the sum of all its terms approaches a specific finite value as more terms are added.
A geometric series is an example where each term is a constant multiple of the previous term. In the case of convergence, for a geometric series to be convergent, the absolute value of the common ratio must be less than 1.
For the given series: \[1 + \frac{1}{6} + \frac{1}{6^2} + \frac{1}{6^3} + \frac{1}{6^4} + \tiny\text{other terms}ormalsize\] the common ratio \(\frac{1}{6}\) is indeed less than 1. Therefore, we can conclude that our series is convergent.
Common Ratio
In a geometric series, the pattern of the terms is determined by the common ratio. The common ratio is calculated by dividing any term in the series by the previous term. For our geometric series: \[1 + \frac{1}{6} + \frac{1}{6^2} + \frac{1}{6^3} + \frac{1}{6^4} + \tiny\text{other terms}ormalsize\] we can see that the term \(\frac{1}{6}\) comes right after 1. Dividing \(\frac{1}{6} \text{by} 1 = \frac{1}{6}\), we find the common ratio is \(\frac{1}{6}\). Understanding the common ratio is key while working with geometric series because it determines both the behavior and the sum of the series.
Sum Formula
For finding the sum of a convergent geometric series, we use a special sum formula. This formula allows us to calculate the total sum of an infinite number of terms in the series. The formula is: \[S = \frac{a}{1 - r}\] where 'S' is the sum of the series, 'a' is the first term, and 'r' is the common ratio.
In our case, the first term 'a' is 1, and the common ratio 'r' is \(\frac{1}{6}\). Substituting these values into the formula, we get:\[\frac{1}{1 - \frac{1}{6}} = \frac{1}{\frac{5}{6}} = \frac{6}{5}\] Thus, the sum of the series is \(\frac{6}{5}\). By using the sum formula, you can easily find the sum of any convergent geometric series once you know the first term and the common ratio.

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Most popular questions from this chapter

Let \(p_{4}(x)\) be the fourth Taylor polynomial of \(f(x)=e^{x}\) at \(x=0 .\) Show that the error in using \(p_{4}(.1)\) as an approximation for \(e^{-1}\) is at most \(2.5 \times 10^{-7} .\) [Hint: Observe that if \(x=.1\) and if \(c\) is a number between 0 and \(.1\), then $$ \left.\left|f^{(5)}(c)\right| \leq f^{(5)}(.1)=e^{.1} \leq e^{1} \leq 3 .\right] $$

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