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Use the test of your choice to determine whether the following series converge. $$\sum_{k=1}^{\infty} \frac{2^{k}}{e^{k}-1}$$

Short Answer

Expert verified
Short Answer: The given series is \(\sum_{k=1}^{\infty} \frac{2^{k}}{e^{k}-1}\). Using the Ratio Test, we found the limit of the ratio of the consecutive terms to be equal to 1. Since the limit is equal to 1, the Ratio Test is inconclusive. Therefore, we cannot determine whether the series converges or diverges using the Ratio Test.

Step by step solution

01

Identify the general term of the series

The general term of the given series is as follows: $$a_k = \frac{2^k}{e^k - 1}$$
02

Calculate the ratio of consecutive terms

Calculate the ratio of the consecutive terms as follows: $$\frac{a_{k+1}}{a_k} = \frac{\frac{2^{k+1}}{e^{k+1} - 1}}{\frac{2^k}{e^k - 1}} = \frac{2^{k+1}(e^k - 1)}{2^k(e^{k+1} -1)}$$
03

Take the limit as \(k\) approaches infinity

Evaluate the limit $$\lim_{k \to \infty} \frac{2^{k+1}(e^k - 1)}{2^k(e^{k+1} -1)}$$ Now divide both the numerator and the denominator by \(2^k\): $$\lim_{k \to \infty} \frac{2(e^k - 1)}{e^k(e^{k+1} -1)}$$
04

Further analyze the limit

To analyze the limit better, we can use L'Hopital's Rule to find the limit of the ratio of the infinite sequences. First, let's create a function based on the series: $$f(k) = \frac{2(e^k - 1)}{e^k(e^{k+1} -1)}$$ Now, take derivatives of both numerator and denominator functions: $$f'(k) = \frac{2 e^k}{e^{k} (e^{k+1}-1) + e^{k+1} (e^k - 1)}$$
05

Apply L'Hopital's Rule and evaluate the limit

Now, we will apply L'Hopital's Rule and evaluate the limit: $$\lim_{k \to \infty} f'(k) = \lim_{k \to \infty} \frac{2 e^k}{e^{k} (e^{k+1}-1) + e^{k+1} (e^k - 1)}$$ As \(k\) approaches infinity, \({e^k}\) will grow infinitely faster than the other terms in the limit. Therefore we can simplify the limit: $$\lim_{k \to \infty} f'(k) = \lim_{k \to \infty} \frac{2}{1+1} = 1$$
06

Determine convergence of the series based on the Ratio Test

According to the Ratio Test, the following conclusions can be made: - If the limit is less than 1, the series converges. - If the limit is greater than 1, the series diverges. - If the limit is equal to 1, the Ratio Test is inconclusive. In our case, the limit is equal to 1, which means the Ratio Test is inconclusive for this series. Thus, we cannot determine convergence or divergence using this method.

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Key Concepts

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

Ratio Test
The Ratio Test is a powerful tool often employed to determine the convergence of an infinite series. It helps to establish whether the sum of terms from a series leads to a finite number.

To apply the Ratio Test, we take the limit of the ratio of consecutive terms in the series, as follows: \( L = \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| \).

Here are the rules for interpreting \( L \):
  • If \( L < 1 \), the series converges absolutely.
  • If \( L > 1 \), the series diverges.
  • If \( L = 1 \), the test is inconclusive.

The Ratio Test is particularly useful when dealing with series that have terms involving powers or factorials, as it simplifies the expression significantly. In the exercise, after calculating the limit, the test showed \( L = 1 \), meaning we couldn't conclusively determine convergence or divergence for the series. This suggests the need for other methods to analyze the series further.
Infinite Series
Infinite series are sums of infinitely many terms, often written as \( \sum_{k=1}^{\infty} a_k \), where \( a_k \) represents the general terms of the series. Understanding whether an infinite series converges is crucial, as convergence indicates that the series sum approaches a finite value.
  • If a series converges, you can find a finite number that it approaches.
  • If a series diverges, its sum grows indefinitely as more terms are added.

Some well-known series types include geometric series, harmonic series, and p-series, each with specific criteria for convergence. Analyzing these series involves approaches like the Ratio Test, Root Test, Integral Test, and Comparison Test, among others.

In the provided exercise, the focus is determining the convergence of the series \( \sum_{k=1}^{\infty} \frac{2^k}{e^k - 1} \). However, with the Ratio Test proving inconclusive, it's important to consider other methods to understand the series behavior better.
L'Hopital's Rule
L'Hopital's Rule is a technique used to find limits of indeterminate forms, particularly these forms: \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \).

The rule is particularly useful in calculus to simplify the evaluation of limits in complicated functions. With L'Hopital's Rule, you differentiate the numerator and the denominator separately, then take the limit again. Here is the formal representation:

If \( \lim_{x \to c} \frac{f(x)}{g(x)} = \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then \( \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \).

It is important to note that L'Hopital's Rule may need to be applied more than once if the result is still indeterminate after the first application. In the exercise, the rule was utilized to assess the limit form in the series to determine convergence, demonstrating its strategic use when tackling complex limits and fractions.

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Most popular questions from this chapter

Consider the alternating series $$ \sum_{k=1}^{\infty}(-1)^{k+1} a_{k}, \text { where } a_{k}=\left\\{\begin{array}{cl} \frac{4}{k+1}, & \text { if } k \text { is odd } \\ \frac{2}{k}, & \text { if } k \text { is even } \end{array}\right. $$ a. Write out the first ten terms of the series, group them in pairs, and show that the even partial sums of the series form the (divergent) harmonic series. b. Show that \(\lim _{k \rightarrow \infty} a_{k}=0\) c. Explain why the series diverges even though the terms of the series approach zero.

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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.

Use Exercise 89 to determine how many terms of each series are needed so that the partial sum is within \(10^{-6}\) of the value of the series (that is, to ensure \(\left|R_{n}\right|<10^{-6}\) ). Functions defined as series Suppose a function \(f\) is defined by the geometric series \(f(x)=\sum_{k=0}^{\infty}(-1)^{k} x^{k}\) a. Evaluate \(f(0), f(0.2), f(0.5), f(1),\) and \(f(1.5),\) if possible. b. What is the domain of \(f ?\)

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