Question

Consider the series \(\sum_{k=2}^{\infty} \frac{1}{k(\ln k)^{p}},\) where \(p\) is a real number. a. Use the Integral Test to determine the values of \(p\) for which this series converges. b. Does this series converge faster for \(p=2\) or \(p=3 ?\) Explain.

Short answer

Additionally, compare the speed of convergence for \(p=2\) and \(p=3\). Answer: The series converges for \(p>1\). The speed of convergence is faster for \(p=3\) than for \(p=2\).

Step-by-step solution

The Integral Test states that if we have a continuous, positive, and decreasing function \(f(x)\), then the series \(\sum_{k=1}^{\infty} f(k)\) converges if the integral \(\int_{1}^{\infty} f(x) \, dx\) converges and diverges if the integral diverges. #Step 2: Define the function and check the conditions of the Integral Test#

To use the Integral Test, we must first define our function, \(f(x)\), which in our case is given by \(f(x) = \frac{1}{x(\ln x)^{p}}\). We can see that \(f(x)\) is continuous for \(x \geq 2\), and it is positive for all values of \(x \geq 2\). To show that it is also decreasing, we can calculate the derivative of \(f(x)\) and show that it is negative. Now we'll find the derivative of the function with respect to \(x\):

Derivative of f(x) #

If \(f(x) = \frac{1}{x(\ln x)^{p}}\), then we can rewrite it as \(f(x) = x^{-1}(\ln x)^{-p}\). To find the derivative, we can use the product rule: \((fg)' = f'g + fg'\). Using the power rule, we get: \(f'(x) = (-1)x^{-2}(\ln x)^{-p}\) and \(g'(x) = (-p)(\ln x)^{-p-1}(\frac{1}{x})\). Multiplying these derivatives, we get: $$ (fg)' = \left[-\frac{1}{x^{2}(\ln x)^{p}}\right] +\left[-\frac{p}{x^{2}(\ln x)^{p+1}}\right] = -\frac{1}{x^{2}(\ln x)^{p}}\left(1+\frac{p}{\ln x}\right). $$ We see that the expression in the brackets, \((1+\frac{p}{\ln x})\), will be positive since \(p \geq 0\) and \(\ln{x} \geq 0\). Therefore, \(f(x)\) is decreasing for \(x \geq 2\). Now, let's compute the integral of \(f(x)\) from 2 to infinity. #Step 3: Compute the integral of f(x) and apply the Integral Test#

We'll find the integral of \(f(x)\): $$\int_{2}^{\infty} \frac{1}{x(\ln x)^{p}} dx$$ Now by substituting \(u=\ln x\), so \(du = \frac{1}{x} dx\), we obtain: $$ \int_{\ln 2}^{\infty} \frac{1}{u^{p}} du = \frac{1}{1-p}\big[ u^{1-p}\big]_{\ln 2}^{\infty} = \frac{1}{1-p}\big[\lim_{u \to \infty} \frac{1}{u^{p-1}} - (\ln2)^{1-p}\big] $$ This integral converges if \(p-1>0\), that is, \(p>1\). So, the series converges for \(p>1\). Now, referring to part (b), we'll compare the convergence speed for \(p=2\) and \(p=3\). #Step 4: Compare the convergence speed for p=2 and p=3#

For the series \(\sum_{k=2}^{\infty} \frac{1}{k(\ln k)^{p}}\), we have already found that it converges for both \(p=2\) and \(p=3\). The terms for these cases are: 1. For \(p=2\): \(\frac{1}{k(\ln k)^2}\) 2. For \(p=3\): \(\frac{1}{k(\ln k)^3}\) As \(p\) increases, the term \(\frac{1}{(k\ln k)^p}\) will decrease faster. Since \(p=3\) is greater than \(p=2\), the convergence for the series will be faster for \(p=3\).

Key concepts

Integral Test

The Integral Test is a fundamental tool in calculus used to determine the convergence of infinite series. To apply this test, we begin by identifying a function, \( f(x) \), which must be continuous, positive, and decreasing for all values of \( x \geq N \) for some fixed number \( N \). The test asserts that if the improper integral \( \int_{N}^{\infty} f(x) \, dx \) converges, then the related series \( \sum_{k=N}^{\infty} f(k) \) also converges. Conversely, if this integral diverges, the series does as well.

In the given exercise, the function \( f(x) = \frac{1}{x(\ln x)^{p}} \) is considered for the interval \( x \geq 2 \). We need to ensure our function meets the conditions set by the integral test. It is positive and continuous for \( x \geq 2 \). To confirm it's decreasing, we calculate its derivative and verify it's negative. With these points satisfied, we can directly apply the Integral Test for analyzing convergence.

The conclusion from the exercise is that the integral of \( \int_{2}^{\infty} \frac{1}{x (\ln x)^{p}} \, dx \) converges when \( p > 1 \), marking the convergence range for our series.

Convergence Criteria

In the context of infinite series, convergence criteria help determine under which conditions a series converges. There are several tests available, but in this exercise, we specifically used the Integral Test. The key is to match the function's behavior with well-known convergent patterns.

For the series \( \sum_{k=2}^{\infty} \frac{1}{k(\ln k)^{p}} \), understanding the behavior of the function \( f(x) = \frac{1}{x(\ln x)^{p}} \) helps identify these criteria.

• The function must be continuous from a specific starting point \( N \), usually where it starts to evaluate.
• The function is positive over the interval \( x \geq 2 \).
• The function should continuously decrease as \( x \) increases, ensuring the terms decrease and the Infinite Series matches the behavior required for convergence.

By ensuring the above conditions and performing the necessary substitutions (like \( u = \ln x \)), we establish that the integral converges for all \( p > 1 \). Thus, this series converges when the exponent \( p \) exceeds 1, providing a clear criterion for convergence.

Comparative Convergence Speed

The speed of convergence is a significant aspect when comparing two convergent series. While both series might converge, their convergence rates could be different. This exercise explores this for \( p=2 \) and \( p=3 \) in the series \( \sum_{k=2}^{\infty} \frac{1}{k(\ln k)^{p}} \).

Convergence speed depends on how quickly the terms of the series approach zero. As the power \( p \) increases, the term \( \frac{1}{k(\ln k)^{p}} \) decreases more swiftly, leading to faster convergence. This is because the denominator increases at a higher rate, making individual terms smaller quicker.

In this exercise, comparing \( p = 2 \) and \( p = 3 \):

• For \( p = 2 \), terms of the series are \( \frac{1}{k(\ln k)^2} \)
• For \( p = 3 \), terms of the series are \( \frac{1}{k(\ln k)^3} \)

The terms become smaller more rapidly for \( p = 3 \) than for \( p = 2 \) due to the additional power. Therefore, the series for \( p = 3 \) converges faster, offering a glimpse into how altering a parameter like \( p \) affects the convergence characteristics of a series.