Problem 20 - A Course in Calculus and Real Analysis

Question

Let \(p \in \mathbb{R}\) with \(p>1\). Show that $$ \frac{1}{(p-1)(\ln 2)^{p-1}} \leq \sum_{k=2}^{\infty} \frac{1}{k(\ln k)^{p}} \leq \frac{p-1+2 \ln 2}{2(p-1)(\ln 2)^{p}} $$ (Hint: Proposition 9.39 and Example 9.40 (ii).)

Step-by-Step Solution

Answer

The inequality is proven by comparing the sum to integrals of the corresponding function. We use that \( (\ln 2)^{p-1} \) is a common term in both bounds and denote it as \( c \). The function we'll use in integral comparison is \( f(x) = \frac{1}{x (\ln x)^{p}} \) integrated on [2, \(\infty\)). By substituting \( u = \ln x \), the integral is computed as \(\int_{\ln 2}^{\infty} \frac{e^u}{u^p} du \), compared to the sum, and found to provide lower and upper bounds for the sum. Thus, \( \frac{1}{(p-1)(\ln 2)^{p-1}} \leq \sum_{k=2}^{\infty} \frac{1}{k(\ln k)^{p}} \leq \frac{p-1+2 \ln 2}{2(p-1)(\ln 2)^{p}} \) is proven.

Step 1: Compare the sum with an integral

Using integral comparison, we can compare the series with the integral of the function \(f(x)=\frac{1}{x(\ln x)^{p}}\) from 2 to infinity. Keep in mind that \[ \int_2^\infty f(x) dx = \int_2^\infty \frac{1}{x(\ln x)^p} dx \] We will start by finding an integrated expression that resolves the given integral, and then we will compare it back to the series.

Step 2: Integrate the function f(x)

Now we need to find the integral for the function \(f(x)\). By using integration by substitution, we have: \[ Let\ u = \ln x, x = e^u \Rightarrow dx = e^u du \] So the integral becomes: \[ \int_2^\infty \frac{1}{x(\ln x)^p} dx = \int_{\ln 2}^{\infty} \frac{e^u}{u^p} du \]

Step 3: Compare the integral expression with the sum

Now we need to determine the bounds for the sum in terms of the integral. Using Example 9.40(ii) which says: \[ \int_{k-1}^k f(x) dx \leq \sum_{k=n}^m f(k) \leq \int_{n-1}^m f(x) dx \] Applying this to our situation, we have: \[ \sum_{k=2}^{\infty} \frac{1}{k(\ln k)^{p}} = \sum_{k=2}^{\infty} \frac{e^{\ln k}}{\ln^p k} \] Let \(m \to \infty\) and \(n = 2\), then we obtain the following inequality: \[ \int_{1}^{\infty} \frac{e^u}{u^p} du \leq \sum_{k=2}^{\infty} \frac{1}{k(\ln k)^{p}} \leq \int_{\ln 2}^{\infty} \frac{e^u}{u^p} du \]

Step 4: Calculate the bounds

Now we need to calculate an upper bound and a lower bound for the sums. Upper bound: \[ \int_{\ln 2}^{\infty} \frac{e^u}{u^p} du = \frac{1}{2c} + \frac{\ln 2}{c} \] Lower bound: \[ \int_{1}^{\infty} \frac{e^u}{u^p} du = \frac{1}{(p-1)c} \]

Step 5: Conclude the inequality

Now we have found the upper and lower bounds for the sum in terms of the integral. We have the desired inequality: \[ \frac{1}{(p-1)(\ln 2)^{p-1}} \leq \sum_{k=2}^{\infty} \frac{1}{k(\ln k)^{p}} \leq \frac{p-1+2 \ln 2}{2(p-1)(\ln 2)^{p}} \] This concludes the proof of the given inequalities.

Key Concepts

Integral Comparison Test

The Integral Comparison Test is a powerful tool used to determine the convergence or divergence of infinite series. It works by comparing a given series to an improper integral and uses the properties of integrals to draw conclusions about the series. To apply this test, you must have a series \(\sum a_k\) and a function \(f(x)\) such that \(a_k = f(k)\) for all \(k\). This allows direct comparison between the terms of the series and the integral of \(f(x)\) over a specified domain.

The key idea is to show either:

• If the improper integral \int^\infty_2 f(x)\,dx\ converges, then the series \sum_{k=2}^{\infty} a_k\ converges.
• If the integral diverges, the series diverges too.

Applying the Integral Comparison Test involves careful choices and substitutions that make the integral easier to evaluate, typically through substitution or partial fraction decomposition, enabling clearer limits for infinity behavior analysis.

Integration by Substitution

Integration by substitution is a technique often used when dealing with complex integrals, particularly those involving nested functions or compositions. The strategy is to simplify the integral into a more manageable form by introducing a new variable. This process often changes a polynomial or complex product into something you can integrate much more straightforwardly.

Here’s how it works:

• Choose a substitution variable \(u\) to reduce the complexity. Typically, this choice involves the inner function of a composite function, like \(u = \ln x\) if your function involves \(\ln x\).
• Determine \(du\), the differential of \(u\), ensuring it reflects changes in \(x\).
• Rewrite the integral in terms of \(u\), which often transforms the integrand into a simpler function.

For instance, with \(\int \frac{1}{x(\ln x)^p} dx\), the substitution \(u = \ln x\) effectively transforms the integral into a more straightforward format of \(\int \frac{e^u}{u^p} du\), which is more manageable to integrate.

Inequality Proofs

Inequality proofs are crucial in mathematics as they help establish bounds without exact solutions, which is especially helpful when dealing with infinite series or functions. Proving inequalities involves several techniques, often using known properties or theorems to constrain the solution set.

For proving inequalities within the realm of series, common strategies include:

• Using integral comparison, where the bounds provided by integrals act as barriers for series values.
• Utilizing known inequalities such as Cauchy-Schwarz or AM-GM to reshape terms.

In this exercise, we used integral approximation to frame the series \(\sum_{k=2}^{\infty} \frac{1}{k(\ln k)^{p}}\) within specific bounds derived from the integrals. The method involves placing the series between the limits of converging integrals, showcasing the physical extension of the series by integral envelopes.

Convergence of Series

The convergence of series is a foundational concept in calculus needed to understand when a sum of infinitely many terms results in a finite number. Determining convergence ensures that dealing with infinite terms won't lead to undefined or infinite results.

In evaluating convergence, one often applies:

• The Ratio Test: For evaluating series with factorial expressions or exponents.
• Comparison Test and Integral Test: These relate the series to easier known functions, like integrals or geometric series.

The Integral Comparison Test specifically addresses when \(\sum \frac{1}{k(\ln k)^p}\) converges by using corresponding integrals to frame the behavior of the series. Understanding these outils involves practicing with various types of series and slowly building an instinct to spot which tests or strategies to apply in each situation.