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

And which converges faster, when \(p=2\) or when \(p=3\)? Answer: The series converges for \(p > 1\). The series converges faster when \(p=3\).

Step-by-step solution

Apply the Integral Test

To apply the Integral Test, we first analyze the function \(f(x) = \frac{1}{x(\ln x)^{p}}\), where \(x\geq2\) and \(p\) is a real number. Observe that the function is positive, continuous, and decreasing on the given interval. To use the Integral Test, we need to evaluate the improper integral, $$\int_{2}^{\infty} \frac{1}{x(\ln x)^{p}}dx$$

Evaluate the improper integral

To evaluate the improper integral, let's first make a substitution. Let \(u = \ln x\), then \(du = \frac{dx}{x}\). The integral transforms to, $$\int_{\ln 2}^{\infty} \frac{1}{u^{p}}du$$ Now, we will evaluate the integral and then take the limit as the upper limit approaches infinity. It's important to note that the value of \(p\) will determine the convergence of the integral: $$\int_{\ln 2}^{\infty} \frac{1}{u^{p}}du = \lim_{t\to\infty} \int_{\ln 2}^{t} \frac{1}{u^{p}}du$$

Calculate the integral

For \(p\neq1\), the antiderivative of \(\frac{1}{u^{p}}\) is: $$\frac{u^{1-p}}{1-p}$$ Apply the Fundamental theorem of calculus, we have: $$\lim_{t\to\infty} \left(\frac{u^{1-p}}{1-p}\Big|_{\ln 2}^{t}\right) = \lim_{t\to\infty} \left( \frac{t^{1-p}}{1-p} - \frac{(\ln 2)^{1-p}}{1-p} \right)$$ Now we have to examine the limit as \(t \rightarrow \infty\).

Examine the limit

To determine when this limit converges, we consider different cases for \(p\): 1. If \(p > 1\), then \(1-p < 0\), making the term \(t^{1-p}\) approach 0 as \(t\to\infty\). So, the limit converges. 2. If \(p < 1\), then \(1-p > 0\), making the term \(t^{1-p}\) approach infinity as \(t\to\infty\). So, the limit diverges. According to the Integral Test, the series converges if the integral converges and vice versa. Therefore, the series converges for \(p > 1\). Now, we move on to the second part of the exercise, where we need to compare the convergence rates for \(p=2\) and \(p=3\).

Compare convergence rates

To compare the convergence rates of the series for \(p=2\) and \(p=3\), we can examine the individual terms of each series. We have: $$\text{For } p=2, \frac{1}{k(\ln k)^2}$$ $$\text{For } p=3, \frac{1}{k(\ln k)^3}$$ Since the term for \(p=3\) is smaller than the term for \(p=2\) for every value of \(k\geq2\), the series converges faster when \(p=3\).