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Chapter 8: Problem 12

Use the Ratio Test to determine whether the following series converge. $$\sum_{k=1}^{\infty} \frac{k^{k}}{2^{k}}$$

Short Answer

Expert verified
Answer: Yes, the series converges.

Step by step solution

01

Write out the expression for the ratio of consecutive terms

We need to find the absolute value the ratio of the (k+1)th term to the kth term: $$\left| \frac{\frac{(k+1)^{k+1}}{2^{k+1}}}{\frac{k^k}{2^k}} \right|$$
02

Simplify the expression

First, observe that the absolute value is not needed, as all values inside are positive. Next, let's simplify the fraction by multiplying the numerator and the denominator, respectively. $$\frac{(k+1)^{k+1}}{2^{k+1}} \cdot \frac{2^k}{k^k} = \frac{(k+1)^{k+1} \cdot 2^k}{2^{k+1} \cdot k^k}$$
03

Simplify the exponent terms

We now observe that there's a common term of \(2^k\) in both the numerator and denominator, which can be canceled out: $$\frac{(k+1)^{k+1}}{2 \cdot k^k}$$
04

Use the Ratio Test by finding the limit as k approaches infinity

To apply the Ratio Test, we need to find the limit of the simplified expression as k approaches infinity: $$\lim_{k \to \infty} \frac{(k+1)^{k+1}}{2 \cdot k^k}$$
05

Use a substitution to help evaluate the limit

Let \(u = k+1\), then change the limit: $$\lim_{u \to \infty} \frac{u^{u}}{2(u-1)^{u-1}}$$
06

Divide both numerator and denominator by \((u-1)^{u-1}\) and evaluate the limit

We can divide both the numerator and denominator by \((u-1)^{u-1}\) to obtain a more manageable expression: $$\lim_{u \to \infty} \frac{u^{u} / (u-1)^{u-1}}{2}$$ Now, we bring the denominator inside the limit and rewrite the expression in terms of k: $$\lim_{k \to \infty} \frac{(k+1)^{k+1} / k^k}{2}$$ With the new expression, we recognize the limit as the definition of the exponential function with base (k+1)/k, which equals \(e\) as k approaches infinity. So we get: $$\frac{e}{2}$$
07

Determine the convergence of the series using the Ratio Test

Since the limit of the simplified expression is \(\frac{e}{2}<1\), it meets the criteria of the Ratio Test for convergence. Thus, the given series converges: $$\sum_{k=1}^{\infty} \frac{k^{k}}{2^{k}}$$ converges by the Ratio Test.

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Key Concepts

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

Convergence of Series
The concept of series convergence is central to understanding many phenomena in mathematics, particularly when dealing with infinite series. A series converges if the sequence of its partial sums approaches a finite limit. In the context of calculus, determining whether a series converges helps us understand its behavior as the number of terms grows infinitely large.

To check the convergence of a series, several tests can be applied, among which the Ratio Test is a popular choice. The Ratio Test involves taking the limit of the absolute value of the ratio of consecutive terms of the series. If this limit is less than 1, the series converges. In our example, the expression \( \sum_{k=1}^{\infty} \frac{k^{k}}{2^{k}} \) was shown to converge using the Ratio Test.

Understanding convergence is crucial because it tells us that as we add more terms, the sum gets closer to a specific value, which can be vital for calculations in various fields of science and engineering.
Infinite Series
An infinite series is a sum of an infinite sequence of terms. The concept starts with a sequence \( a_1, a_2, a_3, \ldots \) and considers the sum \( a_1 + a_2 + a_3 + \ldots \), potentially extending to infinity. Working with infinite series is a cornerstone of calculus and mathematical analysis, allowing us to express numbers, functions, and even processes in concise ways.

Infinite series can either converge or diverge.
  • If a series converges, the sum of its infinite terms approaches a finite number.
  • If a series diverges, the sum keeps increasing or decreasing without bound.

For example, the series \( \sum_{k=1}^{\infty} \frac{k^{k}}{2^{k}} \) was shown to converge. Recognizing whether a series is convergent or divergent guides us in applications where precision is key, ensuring that models and calculations are both accurate and meaningful.
Calculus
Calculus is a branch of mathematics focused on change and motion through concepts such as differentiation and integration. It provides powerful tools for understanding the behavior of functions, especially when dealing with infinite processes like those seen in infinite series.

Within calculus, various tests and methods are used to explore and understand the convergence of series. Techniques like the Ratio Test highlight calculus's role in this exploration by providing criteria to analyze series.

By systematically applying these tests, we can evaluate complex expressions that emerge from calculus problems, such as those in physics, engineering, economics, and beyond.
  • One of the key benefits of calculus is its ability to handle continuous change, making it indispensable in fields that analyze growth patterns, waves, and dynamic systems.
  • Calculus not only aids in understanding series and their convergence but also in finding area under curves and solving differential equations.
Ultimately, mastery of calculus enriches our understanding of a wide array of mathematical and real-world phenomena, laying a foundation for advanced theoretical and applied mathematics.

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Most popular questions from this chapter

In Section \(8.3,\) we established that the geometric series \(\sum r^{k}\) converges provided \(|r| < 1\). Notice that if \(-1 < r<0,\) the geometric series is also an alternating series. Use the Alternating Series Test to show that for \(-1 < r <0\), the series \(\sum r^{k}\) converges.

a. Sketch the function \(f(x)=1 / x\) on the interval \([1, n+1]\) where \(n\) is a positive integer. Use this graph to verify that $$\ln (n+1)<1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n}<1+\ln n.$$ b. Let \(S_{n}\) be the sum of the first \(n\) terms of the harmonic series, so part (a) says \(\ln (n+1)0,\) for \(n=1,2,3, \ldots\) c. Using a figure similar to that used in part (a), show that $$\frac{1}{n+1}>\ln (n+2)-\ln (n+1).$$ d. Use parts (a) and (c) to show that \(\left\\{E_{n}\right\\}\) is an increasing sequence \(\left(E_{n+1}>E_{n}\right)\). e. Use part (a) to show that \(\left\\{E_{n}\right\\}\) is bounded above by 1 . f. Conclude from parts (d) and (e) that \(\left\\{E_{n}\right\\}\) has a limit less than or equal to \(1 .\) This limit is known as Euler's constant and is denoted \(\gamma\) (the Greek lowercase gamma). g. By computing terms of \(\left\\{E_{n}\right\\}\), estimate the value of \(\gamma\) and compare it to the value \(\gamma \approx 0.5772 .\) (It has been conjectured that \(\gamma\) is irrational.) h. The preceding arguments show that the sum of the first \(n\) terms of the harmonic series satisfy \(S_{n} \approx 0.5772+\ln (n+1)\) How many terms must be summed for the sum to exceed \(10 ?\)

Evaluate the limit of the following sequences or state that the limit does not exist. $$a_{n}=\int_{1}^{n} x^{-2} d x$$

The expression $$1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+}}}}.$$ where the process continues indefinitely, is called a continued fraction. a. Show that this expression can be built in steps using the recurrence relation \(a_{0}=1, a_{n+1}=1+1 / a_{n},\) for \(n=0,1,2,3, \ldots . .\) Explain why the value of the expression can be interpreted as \(\lim _{n \rightarrow \infty} a_{n},\) provided the limit exists. b. Evaluate the first five terms of the sequence \(\left\\{a_{n}\right\\}\). c. Using computation and/or graphing, estimate the limit of the sequence. d. Assuming the limit exists, use the method of Example 5 to determine the limit exactly. Compare your estimate with \((1+\sqrt{5}) / 2,\) a number known as the golden mean. e. Assuming the limit exists, use the same ideas to determine the value of $$a+\frac{b}{a+\frac{b}{a+\frac{b}{a+\frac{b}{a+}}}}$$ where \(a\) and \(b\) are positive real numbers.

Stirling's formula Complete the following steps to find the values of \(p>0\) for which the series \(\sum_{k=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots(2 k-1)}{p^{k} k !}\) converges. a. Use the Ratio Test to show that \(\sum_{k=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots(2 k-1)}{p^{k} k !}\) converges for \(p>2\). b. Use Stirling's formula, \(k !=\sqrt{2 \pi k} k^{k} e^{-k}\) for large \(k,\) to determine whether the series converges when \(p=2\). (Hint: \(1 \cdot 3 \cdot 5 \cdots(2 k-1)=\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdots(2 k-1) 2 k}{2 \cdot 4 \cdot 6 \cdots 2 k}\) (See the Guided Project Stirling's formula and \(n\) ? for more on this topic.)
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