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Use the test of your choice to determine whether the following series converge. $$\sum_{k=2}^{\infty} \frac{5 \ln k}{k}$$

Short Answer

Expert verified
Based on the integral test and the evaluation of the improper integral, we concluded that the given series $$\sum_{k=2}^{\infty} \frac{5 \ln k}{k}$$ diverges.

Step by step solution

01

Confirm Conditions for Integral Test

First, let's verify that $$f(x) = \frac{5\ln x}{x}$$ is continuous, positive and decreasing on the interval $$[2,\infty)$$. - Continuous: The function is continuous for $$x\in(0,\infty)$$, including $$[2,\infty)$$. - Positive: The function is positive for $$x > 1$$ (since the natural logarithm of 1 is 0). - Decreasing: The derivative of the function is $$f'(x) = \frac{-5}{x^2} \left( \ln x + 1 \right)$$. Since $$x>1$$, the terms inside the brackets are positive, so the derivative is negative, meaning the function is decreasing. Now that we confirmed these conditions, we can apply the integral test.
02

Evaluate the Integral

We have the following improper integral to evaluate: $$\int_2^{\infty} \frac{5\ln x}{x} \, dx$$ To evaluate this integral, we can use integration by parts with $$u=\ln x$$ and $$dv=\frac{5}{x} \,dx$$. Then, we find: $$du = \frac{1}{x} \, dx, \quad v= 5\ln x$$ So, the integration by parts formula gives: $$\int_2^{\infty} \frac{5\ln x}{x} \, dx = \left[ 5\ln x \cdot (\ln x - 1) \right]_2^\infty - \int_2^\infty 5\left( \ln x - 1 \right) \, dx$$ Now we need to evaluate the resulting terms: $$\lim_{t \to \infty} \left[ 5\ln x \cdot (\ln x - 1) \right]_2^t - \int_2^\infty 5\left( \ln x - 1 \right) \, dx$$
03

Evaluate the Limit and the Integral

The first term is: $$\lim_{t \to \infty} \left[ 5\ln x \cdot (\ln x - 1) \right]_2^t = \lim_{t \to \infty} 5\ln t (\ln t - 1) - 5\ln 2(\ln 2 - 1)$$ Since the term $$\ln t (\ln t - 1)$$ grows without bound as $$t$$ goes to infinity, so the limit is infinite. As the limit is infinite, we don't need to evaluate the integral $$\int_2^\infty 5\left( \ln x - 1 \right) \, dx$$, as the expression is already divergent.
04

Conclusion

Since the limit and integral don't converge, we conclude that the given series $$\sum_{k=2}^{\infty} \frac{5 \ln k}{k}$$ also diverges according to the integral test.

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Key Concepts

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

Integral Test
The Integral Test is a useful method for determining the convergence or divergence of an infinite series. It revolves around comparing a series to an improper integral. This method is applicable when you have a function \( f(x) \) that is continuous, positive, and decreasing for all \( x \geq N \), where \( N \) is a positive integer.

The steps to apply the Integral Test are simple:
  • First, ensure the function \( f(x) \) is continuous, positive, and decreasing.
  • Second, set up the improper integral \( \int_{N}^{\infty} f(x) \, dx \).
  • Evaluate this integral. If it converges, the series \( \sum_{k=N}^{\infty} a_k \) converges as well. Conversely, if the integral diverges, the series also diverges.


Applying these steps to a series provides a strong tool, especially for understanding how the terms grow or shrink across large indices. The key alignment between the integral and the sum of the series forms the basis of this powerful test.
Improper Integrals
Improper integrals play an essential role in calculus, especially when dealing with unbounded intervals or functions that have vertical asymptotes. They essentially allow you to calculate areas under curves that extend indefinitely. When you encounter notation like \( \int_{2}^{\infty} \frac{5 \ln x}{x} \, dx \), it signals an improper integral.

There are two key types of improper integrals:
  • Type I: Integrals with an infinite limit of integration, such as \( \int_{a}^{\infty} f(x) \, dx \).
  • Type II: Integrals with a function that possesses a vertical asymptote within the integration interval, such as \( \int_{a}^{b} \frac{1}{(c-x)^p} \, dx \).

In either case, evaluating these integrals often involves taking limits. You explore limits as the variable approaches infinity or the point of discontinuity. The convergence of these integrals significantly impacts the convergence of related series in problems such as those involving the Integral Test.
Logarithmic Functions
Logarithmic functions, often denoted as \( \ln x \) for natural logarithms, are vital in calculus and analysis, often appearing in series and integrals. They inherently grow slower than power functions but are instrumental in mathematical analysis.

When working with logarithms:
  • Understand their domain: \( \ln x \) is defined for \( x > 0 \) and shows how slow growth influences functions such as \( \frac{\ln x}{x} \).
  • Recognize that logarithms are continuous everywhere they are defined, making them ideal for integral manipulation.
  • Acknowledge their properties: Logarithm identities like \( \ln(ab) = \ln a + \ln b \) or \( \ln a^b = b \ln a \) are pivotal during problem-solving.

Their properties also ensure logarithms impact convergence tests, which are often part of the complexity when using the Integral Test. In essence, recognizing how the logarithmic function behaves supports the calculation of integrals and helps predict the behavior of sequences and series.

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Most popular questions from this chapter

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