Problem 1 - The Calculus with Analytic Geometry

Question

Determine if the given series is convergent or divergent.\(\sum_{n=1}^{+\infty} \frac{1}{n 2^{n}}\)

Step-by-Step Solution

Answer

The series \( \sum_{n=1}^{+\infty} \frac{\ln n}{n} \) is divergent.

Step 1: Identify the Series

The given series is \(\sum_{n=1}^{+\infty} \frac{\ln n}{n}\). We need to determine if this series is convergent or divergent.

Step 2: Apply the Integral Test

To apply the integral test, consider the function \(f(x) = \frac{\ln x}{x}\). Verify if \(f(x)\) is positive, continuous, and decreasing for \(x \ge 1\). Since \(\ln x\) increases slower than \(x\), the function is indeed decreasing for \(x\ge e\). Thus, we can apply the integral test.

Step 3: Compute the Improper Integral

Evaluate the improper integral \(\int_{1}^{+\infty} \frac{\ln x}{x} \, dx \). Use integration by parts: let \(u = \ln x\) and \(dv = \frac{1}{x} dx\). Then \(du = \frac{1}{x} dx\) and \(v = \ln x\), so the integral becomes \(\int_{1}^{+\infty} \frac{\ln x}{x} \, dx = \left. \ln x \cdot \ln x \right|_{1}^{+\infty} - \int_{1}^{+\infty} \frac{1}{x} \, dx \).

Step 4: Simplify the Integral

The boundary term \( \left. \ln x \cdot \ln x \right|_{1}^{+\infty} \) diverges because \( \ln x \) tends to infinity as \( x \to +\infty \). The remaining integral of \( \frac{1}{x} \) also diverges. Thus, the integral \( \int_{1}^{+\infty} \frac{\ln x}{x} \, dx \) diverges.

Step 5: Conclusion from the Integral Test

Since the improper integral \( \int_{1}^{+\infty} \frac{\ln x}{x} \, dx \) diverges, by the integral test, the given series \( \sum_{n=1}^{+\infty} \frac{\ln n}{n} \) is also divergent.

Key Concepts

integral test

The integral test is an essential tool for determining if a series converges or diverges. For the integral test, we compare the sum of a series to the integral of a related function. The test involves three main steps:

• First, define a function that matches the series' terms. For example, if the series is \(\frac{1}{n^p}\), use the function \(f(x) = \frac{1}{x^p}\).
• Verify that the function is positive, continuous, and decreasing. This ensures the behavior of the integral is similar to the series.
• Evaluate the improper integral from 1 to infinity. If the integral converges, the series converges. If the integral diverges, the series diverges.

In our example, we used the function \(f(x) = \frac{\text{ln} x}{x}\). By calculating the integral and noting it diverges, we concluded that the original series \( \frac{\text{ln} n}{n}\) is divergent.

improper integral

Improper integrals extend the concept of integrals to unbounded intervals or integrands with unbounded behavior at some points. When solving improper integrals:

• Evaluate the integral's behavior at the boundaries. If the boundary is infinity, consider the limit as it approaches infinity.
• If the integral has a discontinuity or an unbounded limit, split it at the problem point and evaluate the limits separately.

For example, to compute \( \frac{\text{ln} x}{x}\) from 1 to infinity, we set up the integral \( \text{\textbackslash}int_{1}^{+\text{infinity}} \frac{\text{ln} x}{x} dx\). By evaluating this, we find that both the limit and remaining integral diverge, leading to a divergent series.

comparison test

The comparison test is another method to determine the convergence/divergence of a series.

• First, identify a simpler series that is similar to the given series and has known convergence properties.
• Compare the terms of the two series: if the given series is smaller term-by-term than a known convergent series, it will also converge. Conversely, if it is larger than a divergent series, it will also diverge.

In practice with our example series \(\frac{\text{ln} n}{n}\), we can compare it to the series \(\frac{1}{n}\), also known as the harmonic series, which diverges. Since \(\frac{\text{ln} n}{n}\) grows slower, the comparison test would similarly show its divergence.

integration by parts

Integration by parts is a technique that applies the product rule of differentiation in reverse. It's useful for integrals involving the product of functions. The formula is: \[ \text{\textbackslash}int u \frac{d}{dx}v \,dx = uv - \text{\textbackslash}int v \frac{d}{dx}u \,dx \] Here's how you solve it:

• Select \(u\) and \(dv\) wisely to simplify the integral.
• Differentiate \(u\) to get \(du\) and integrate \(dv\) to get \(v\).
• Substitute these into the formula.

For our problem, we chose \(u = \text{ln} x\) and \(dv = \frac{1}{x} dx\). After differentiating and integrating, substituting back simplified our integral, showing both parts diverge.