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Chapter 8: Problem 43

Use the properties of infinite series to evaluate the following series. $$\sum_{k=1}^{\infty} \frac{4}{12^{k}}$$

Short Answer

Expert verified
$$\sum_{k=1}^{\infty} \frac{4}{12^{k}}$$ Answer: The sum of the given infinite geometric series is $$\frac{4}{11}$$.

Step by step solution

01

Identify the first term and the common ratio

For the given geometric series $$\sum_{k=1}^{\infty} \frac{4}{12^{k}}$$, the first term is obtained when k = 1, which is $$\frac{4}{12} = \frac{1}{3}$$. The common ratio of the series can be determined by dividing any term by the previous term. Let's divide the term with k = 2 by the term with k = 1: $$\frac{\frac{4}{12^2}}{\frac{4}{12}} = \frac{1}{12}$$ Thus, the common ratio is $$\frac{1}{12}$$.
02

Check if the series converges

A geometric series converges if the common ratio, |r|, is between -1 and 1, which means -1 < |r| < 1. The common ratio in this case is $$\frac{1}{12}$$ and its absolute value |r| = $$\frac{1}{12}$$, which means the series converges.
03

Calculate the sum of the infinite series

Since the given series is convergent, we can use the formula for the sum of an infinite geometric series to find its value: $$S = \frac{a}{1 - r}$$ where S is the sum of the series, a is the first term, and r is the common ratio. In our case, a = $$\frac{1}{3}$$ and r = $$\frac{1}{12}$$. Plugging these values into the formula, we get: $$S = \frac{\frac{1}{3}}{1 - \frac{1}{12}} = \frac{\frac{1}{3}}{\frac{11}{12}}$$ To calculate the sum, we can multiply both the numerator and the denominator of the fraction by 12: $$S = \frac{\frac{4}{12}}{\frac{11}{1}}$$ Now, we can simplify the fraction to find the sum of the series: $$S = \frac{4}{11}$$ Thus, the sum of the given infinite series is $$\frac{4}{11}$$.

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Key Concepts

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

Convergence of Series
In understanding infinite series, particularly the geometric series, convergence is a key concept. A series converges when the sum of its infinite terms approaches a finite number. This happens if the absolute value of the series' common ratio is less than one:
  • For a geometric series \( \sum_{k=1}^{\infty} ar^k \), where \( a \)is the first term and \( r \) is the common ratio, convergence is guaranteed when \( |r| < 1 \).
  • Essentially, convergence implies that adding more and more terms will lead us closer to a specific number rather than leading to infinity.
In the problem given, the common ratio was \( rac{1}{12} \). Since \( rac{1}{12} \) is much less than one, the series converges
  • This means that even if we add an infinite number of terms, the sum will stabilize at a particular value rather than shooting off to infinity.
Understanding convergence is crucial because it allows us to work with infinite series as if they were finite sums, greatly simplifying mathematical calculations.
Infinite Series
An infinite series is essentially an endless sum, with an infinite number of terms adding up. For geometric series, each term is obtained by multiplying the previous term by a constant called the common ratio. Infinite series play a vital role in numerous practical applications, from computing probabilities to solving differential equations.
  • Geometric series are a huge subset of infinite series where each term鈥檚 ratio to the previous one remains constant, defined mathematically as \sum_{k=0}^{\infty} ar^k, where \( a \) is the first term, and \( r \) is the common ratio.
  • It鈥檚 fascinating that despite the infinite nature of these series, many geometric series can produce a finite sum.
In the example provided in the exercise, the first term was \( rac{1}{3} \), and the common ratio was \( rac{1}{12} \). Developing an understanding of infinite series helps in grasping more complex mathematical concepts and provides tools for modeling and solving real-world problems.
Series Sum Formula
The series sum formula for a convergent geometric series allows us to find the sum of an infinite series quickly and efficiently. The formula is expressed as:\[S = \frac{a}{1 - r} \]where
  • \( S \) is the sum of the series
  • \( a \) is the first term
  • \( r \) is the common ratio
For the series in the exercise, \( a = \frac{1}{3} \) and \( r = \frac{1}{12} \). Plugging these values into the series sum formula we get:
  • Converting this gives us \( S = \frac{\frac{1}{3}}{1- \frac{1}{12}} \).
Further simplifying yields \( S = \frac{\frac{1}{3}}{\frac{11}{12}} \).Upon simplification, the sum \( S = \frac{4}{11} \), giving us the total for the series.
  • This remarkable tool lets you discover just how manageable infinite series can become with the right mathematical tools and approaches.
The series sum formula empowers mathematical exploration by distilling infinite complexity into understandable parts.

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Most popular questions from this chapter

Use the formal definition of the limit of a sequence to prove the following limits. $$\lim _{n \rightarrow \infty} b^{-n}=0, \text { for } b > 1$$

In \(1978,\) in an effort to reduce population growth, China instituted a policy that allows only one child per family. One unintended consequence has been that, because of a cultural bias toward sons, China now has many more young boys than girls. To solve this problem, some people have suggested replacing the one-child policy with a one-son policy: A family may have children until a boy is born. Suppose that the one-son policy were implemented and that natural birth rates remained the same (half boys and half girls). Using geometric series, compare the total number of children under the two policies.

In the following exercises, two sequences are given, one of which initially has smaller values, but eventually "overtakes" the other sequence. Find the sequence with the larger growth rate and the value of \(n\) at which it overtakes the other sequence. $$a_{n}=n^{1.001} \text { and } b_{n}=\ln n^{10}, n \geq 1$$

Evaluate the limit of the following sequences or state that the limit does not exist. $$a_{n}=\frac{6^{n}+3^{n}}{6^{n}+n^{100}}$$

An early limit Working in the early 1600 s, the mathematicians Wallis, Pascal, and Fermat were calculating the area of the region under the curve \(y=x^{p}\) between \(x=0\) and \(x=1,\) where \(p\) is a positive integer. Using arguments that predated the Fundamental Theorem of Calculus, they were able to prove that $$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=0}^{n-1}\left(\frac{k}{n}\right)^{p}=\frac{1}{p+1}$$ Use what you know about Riemann sums and integrals to verify this limit.
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