Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 11: Problem 22
Determine whether the series is convergent or divergent. $$ \sum_{n=2}^{\infty} \frac{\ln n}{n^{2}} $$
Short Answer
Expert verified
The series is convergent.
Step by step solution
01
Identify the Series Type
We have the series \( \sum_{n=2}^{\infty} \frac{\ln n}{n^{2}} \). This is an infinite series with terms of the form \( a_n = \frac{\ln n}{n^2} \). The series has a form similar to a \( p\)-series, but with a logarithmic adjustment.
02
Compare with Known Convergent Series
The standard \( p\)-series \( \sum \frac{1}{n^p} \) converges if \( p > 1 \). Here, our term \( \ln n \) grows slower than any polynomial, whereas \( \frac{1}{n^2} \) represents a convergent \( p\)-series with \( p = 2 \). Hence, we suspect the given series is convergent because the logarithmic term will not significantly affect the convergence provided by \( n^{-2} \).
03
Utilize the Integral Test for Convergence
To determine convergence, apply the integral test. Consider the function \( f(x) = \frac{\ln x}{x^2} \) that is continuous, positive, and decreasing on \([2, \infty)\). Evaluate the improper integral:\[\int_{2}^{\infty} \frac{\ln x}{x^2} \, dx.\]Perform integration by parts, where \( u = \ln x \) and \( dv = \frac{1}{x^2} \, dx \). Then \( du = \frac{1}{x} \, dx \) and \( v = -\frac{1}{x} \). The integral becomes:\[-\frac{\ln x}{x} \bigg|_2^{\infty} + \int_{2}^{\infty} \frac{1}{x^2} \, dx.\]The first term evaluates to 0 as \( x \to \infty \) and \(-\ln 2/2\) at 2, and the second integral \( \int_{2}^{\infty} \frac{1}{x^2} \, dx = \left. -\frac{1}{x} \right|_2^{\infty} = \frac{1}{2} \) is convergent. Both terms result in a finite value.
04
Conclusion
Since the integral \( \int_{2}^{\infty} \frac{\ln x}{x^2} \, dx \) converges, by the integral test, the series \( \sum_{n=2}^{\infty} \frac{\ln n}{n^2} \) is also convergent.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integral Test
The Integral Test is a tool used to determine the convergence or divergence of infinite series. It links the behavior of an infinite series to an improper integral. If the terms of a series \((a_n)\) are positive, continuous, and decreasing, the series can be closely related to a function \(f(x)\):
\[a_n = f(x).\]
For the series \(\sum_{n=2}^{\infty}a_n\), if the improper integral
\[\int_{2}^{\infty} f(x) \, dx\]
converges, then the series converges. Conversely, if the integral diverges, the series diverges. This method helps in situations where comparison with a \(p\)-series or direct analysis is complicated.
Here, applying the Integral Test to the series \(\sum_{n=2}^{\infty} \frac{\ln n}{n^2}\), we found the associated continuous function \(f(x) = \frac{\ln x}{x^2}\). Evaluating its integral from 2 to infinity, the integral converged, confirming the series' convergence.
\[a_n = f(x).\]
For the series \(\sum_{n=2}^{\infty}a_n\), if the improper integral
\[\int_{2}^{\infty} f(x) \, dx\]
converges, then the series converges. Conversely, if the integral diverges, the series diverges. This method helps in situations where comparison with a \(p\)-series or direct analysis is complicated.
Here, applying the Integral Test to the series \(\sum_{n=2}^{\infty} \frac{\ln n}{n^2}\), we found the associated continuous function \(f(x) = \frac{\ln x}{x^2}\). Evaluating its integral from 2 to infinity, the integral converged, confirming the series' convergence.
Improper Integrals
Improper integrals help in evaluating areas under curves extending indefinitely, either in terms of the function having an infinite range or having an unbounded region. These integrals are crucial when applying the Integral Test for determining series convergence.
In our situation, the improper integral:
\[\int_{2}^{\infty} \frac{\ln x}{x^2} \, dx \]
was assessed using integration by parts. Key steps included choosing components: \(u = \ln x\) with \(dv = \frac{1}{x^2} \, dx\). Through these calculations, we observed the importance of stepping through the mechanics of integration by parts.
The result of solving this integral was a finite value, indicating the convergence of the integral. This convergence translated to confirming the convergence of the given series.
In our situation, the improper integral:
\[\int_{2}^{\infty} \frac{\ln x}{x^2} \, dx \]
was assessed using integration by parts. Key steps included choosing components: \(u = \ln x\) with \(dv = \frac{1}{x^2} \, dx\). Through these calculations, we observed the importance of stepping through the mechanics of integration by parts.
The result of solving this integral was a finite value, indicating the convergence of the integral. This convergence translated to confirming the convergence of the given series.
Logarithmic Functions
Logarithmic functions, such as \(\ln x\), are vital in mathematical analysis, often included in series terms. In the given series, the term \(\ln n\) modifies the convergence behavior slightly. Though slower growing compared to polynomials and often deemed negligible compared to powers like \(n^{-2}\), the logarithm warrants special attention during analysis.
The primary properties of \(\ln x\) relevant here include:
The primary properties of \(\ln x\) relevant here include:
- Its slow rate of growth compared to polynomial functions.
- A smooth, continuous curve appropriate for integration techniques.
- Its behavior as \(x\) becomes large, helping determine long-term series behavior.
