Question

The prime numbers are those positive integers that are divisible by only 1 and themselves (for example, 2,3,5,7, 11,13, \(\ldots\) ). A celebrated theorem states that the sequence of prime numbers \(\left\\{p_{k}\right\\}\) satisfies \(\lim _{k \rightarrow \infty} p_{k} /(k \ln k)=1 .\) Show that \(\sum_{k=2}^{\infty} \frac{1}{k \ln k}\) diverges, which implies that the series \(\sum_{k=1}^{\infty} \frac{1}{p_{k}}\) diverges.

Short answer

Question: Prove that the series \(\sum_{k=2}^{\infty} \frac{1}{k \ln k}\) diverges, and explain what this implies about the series of the reciprocals of prime numbers. Answer: Using the Integral Test, we showed that the function \(f(x) = \frac{1}{x \ln x}\) is positive, continuous, and decreasing on the interval \([2, \infty)\). We then evaluated the integral of this function from \(2\) to \(\infty\), which was found to be divergent. Because the integral diverged, the corresponding series \(\sum_{k=2}^{\infty} \frac{1}{k \ln k}\) also diverges. This implies that the series of the reciprocals of prime numbers \(\sum_{k=1}^{\infty} \frac{1}{p_{k}}\) diverges, since each term in this series is a term of the divergent \(\sum_{k=2}^{\infty} \frac{1}{k \ln k}\) series.

Step-by-step solution

Define the function

Let's define the function \(f(x) = \frac{1}{x \ln x}\) on the interval \([2, \infty)\).

Verify the conditions for the Integral Test

First, we need to verify that \(f(x)\) is positive, continuous, and decreasing on the interval \([2, \infty)\): 1. \(f(x) = \frac{1}{x \ln x}\) is positive in our interval because both \(x\) and \(\ln x\) are positive for \(x \ge 2\). 2. \(f(x)\) is continuous as both \(x\) and \(\ln x\) are continuous in our interval and their product is nonzero. 3. To prove \(f(x)\) is decreasing, we can show that its derivative is negative. We'll compute the derivative of \(f(x)\) and verify its sign.

Compute the derivative of f(x)

Apply the quotient rule to find the derivative of \(f(x)\): \(f'(x) = \frac{-1}{(x \ln x)^2} \cdot (\ln x + 1)\). As we can see, \((x\ln x)^2\) is always positive in the interval \([2, \infty)\). The term \((\ln x + 1)\) is positive as well, as \(\ln 2 > 0\). Therefore, \(f'(x) < 0\). So, f(x) is decreasing in the interval \([2, \infty)\).

Evaluate the integral

Now we apply the Integral Test. We'll evaluate the integral of the function \(f(x)\) from \(2\) to \(\infty\): $$ \int_{2}^{\infty} f(x) dx = \int_{2}^{\infty} \frac{1}{x \ln x} dx. $$ Let's use the substitution, \(u =\ln x \Rightarrow du = \frac{1}{x} dx\). The integral becomes: $$ \int_{\ln 2}^{\infty} \frac{1}{u} du = \lim_{t \to \infty} \int_{\ln 2}^{t} \frac{1}{u} du $$

Evaluate the limit of the integral

Calculate the antiderivative of \(\frac{1}{u}\) and evaluate the limit: $$ \lim_{t \to \infty} \left[ \ln u \right]_{\ln 2}^{t} = \lim_{t \to \infty} (\ln t - \ln(\ln 2)). $$ This limit is infinite as \(t\) goes to infinity. Therefore, the integral diverges.

Final conclusion

Since \(\int_{2}^{\infty} \frac{1}{x \ln x} dx\) diverges, by the Integral Test we conclude that the series \(\sum_{k=2}^{\infty} \frac{1}{k \ln k}\) diverges as well. This implies that the series of the reciprocals of prime numbers \(\sum_{k=1}^{\infty} \frac{1}{p_{k}}\) also diverges, because each term in this series is a term of the divergent \(\sum_{k=2}^{\infty} \frac{1}{k \ln k}\) series.

Key concepts

Integral Test

The Integral Test is an essential tool in calculus used to determine the convergence or divergence of infinite series. This test applies specifically to series of the form \( \sum_{k=1}^{\infty} a_k \), where \( a_k \) corresponds to a function \( f(x) \) that is positive, continuous, and decreasing from some point onward.

Here's how it works:

• Given \( f(x) \), verify it meets the criteria above on a given interval.
• Evaluate the improper integral \( \int_a^{\infty} f(x) \, dx \).

If the integral converges, the series converges. If the integral diverges, the series also diverges. In our problem, \( f(x) = \frac{1}{x \ln x} \) is positive, continuous, and decreases for \( x \geq 2 \). By evaluating \( \int_2^{\infty} \frac{1}{x \ln x} \, dx \), and finding it diverges, we deduce the series \( \sum_{k=2}^{\infty} \frac{1}{k \ln k} \) diverges as well. This is the crux of the Integral Test.

Prime Numbers

Prime numbers are fascinating and fundamental components of number theory. A prime number is a positive integer greater than 1, divisible only by 1 and itself. Common examples include 2, 3, 5, 7, 11, etc. Unlike composite numbers, primes cannot be broken into smaller multiplicative components.

In the realm of advanced mathematics:

• The distribution of prime numbers reveals intriguing patterns, and understanding these patterns helps us decipher the nature of numbers.
• The Euler prime product formula connects prime numbers to logarithmic functions, illustrating the deep connections across different branches of mathematics.

Our exercise revolves around how the reciprocal series of prime numbers diverges, shedding light on the density and distribution of primes.

Divergence of Series

The concept of divergence is pivotal in understanding infinite series. A series \( \sum_{k=1}^{\infty} a_k \) diverges if the sum cannot be pinned down to a finite number as \( k \) approaches infinity.

Key characteristics of divergent series include:

• Despite having terms that approach zero, the total sum grows indefinitely.
• Behavior that's often clarified using tests like the Integral Test or Comparison Test.

In the exercise, proving that \( \sum_{k=2}^{\infty} \frac{1}{k \ln k} \) diverges is crucial. This divergence insightfully implies that the series \( \sum_{k=1}^{\infty} \frac{1}{p_k} \) concerning primes also diverges, meaning that the sum of reciprocals of primes grows without bound.

Calculus

Calculus is a cornerstone of modern mathematics, dealing with change and motion, as well as the infinite. The study of calculus encompasses two major branches: differentiation and integration.

Differentiation involves finding rates of change and slopes of curves, while integration deals with areas under curves and accumulation of quantities. These concepts are crucial when evaluating the behavior of functions over intervals, especially when examining divergent or convergent series.

In our exercise, calculus, specifically integration, plays a vital role. By evaluating the integral \( \int_2^{\infty} \frac{1}{x \ln x} \, dx \), calculus helps verify the divergence of this particular integral, leading to conclusions about the related series. Understanding these core calculus concepts allows students to tackle complex series problems with greater clarity.