Question

Does \(\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{p}}\) converge if \(p\) is large enough? If so, for which \(p ?\)

Short answer

The series converges for \( p > 1 \).

Step-by-step solution

Identify the series

The series given is \(\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{p}}\). We need to determine if this series converges, and if so, find which values of \(p\) allow it to converge.

Consider the test for convergence

The series \(\sum_{n=2}^{\infty} \frac{1}{n (\ln n)^p}\) has a form similar to a generalized \(p\)-series. This suggests that using the Integral Test for convergence could be appropriate.

Set up the integral for the Integral Test

Use the Integral Test: Consider the integral \(\int_{2}^{\infty} \frac{1}{x (\ln x)^p} \, dx\). If this integral converges, then the series converges.

Perform substitution for easier integration

Let \( u = \ln x \), therefore \( du = \frac{1}{x} \, dx \), and the integral becomes \(\int \frac{1}{u^p} \, du\).

Evaluate the integral

The integral \(\int \frac{1}{u^p} \, du\) gives \(\frac{u^{1-p}}{1-p}\) for \(p eq 1\). Evaluate from \( u = \ln 2 \) to infinity.

Analyze the convergence from the integral result

For \( p > 1 \), the integral \(\int \frac{1}{u^p} \, du\) converges because as \( u \to \infty \), the term \(\frac{u^{1-p}}{1-p}\) approaches 0. For \( p \leq 1 \), the integral diverges.

Key concepts

Integral Test

The Integral Test is a crucial tool in determining the convergence of a series. This test is particularly useful for series that resemble integrable functions. Here’s how it works: if you have a series \(\sum_{n=k}^{\infty} a_n\) where \(a_n\) is positive, continuous, and decreasing, you can compare it to the improper integral \(\int_{k}^{\infty} f(x) \, dx\).

• If the integral converges and is finite, the series will also converge.
• If the integral diverges, the series will also diverge.

For our problem, this test aligns perfectly because we're dealing with a series expressed in an integrable format: \(\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^p}\). By transitioning this series into an integral, we pave the way to determine convergence by evaluating the integral’s behavior.

Generalized p-series

A generalized \( p \)-series is a broad category of series that takes the form \(\sum \frac{1}{n^q}\), where \(q > 0\). The convergence of such series heavily relies on the value of \(q\).

• If \(q > 1\), the series converges.
• If \(q \leq 1\), the series diverges.

Our exercise presents a slightly altered structure with logarithmic functions involved, specifically in the form \(\sum \frac{1}{n(\ln n)^p}\). Despite the complexity added by the logarithm, the behavior is still influenced by \( p \) and mimics the logical framework of a \( p \)-series when assessed through the Integral Test. The goal is to identify a threshold for \( p \) that ensures convergence, similar to how \( q \) affects a \( p \)-series.

Convergence Criteria

Understanding when a series converges is essential to solving problems like these. The convergence criteria for our specific series arise from the Integral Test applied to the transformed function.
If the integral \(\int_{2}^{\infty} \frac{1}{x (\ln x)^p} \, dx\) is examined, convergence depends on the value of \( p \) in relation to the integrand’s behavior as \( x \to \infty \).

• For \( p > 1 \), the expression in the integral decreases rapidly enough to converge to 0, ensuring the integral converges.
• Conversely, if \( p \leq 1 \), the integral diverges because the integrand does not shrink quickly enough to allow the area under the curve to remain finite.

These criteria are derived directly from evaluating the resulting transformed integral, guiding us in determining for which values of \( p \) the series initially provided will converge.

Substitution Method

The Substitution Method is an effective strategy in calculus for simplifying integrals, especially when facing complex expressions. This technique involves changing variables to make the integration process more manageable. In our problem, the substitution \( u = \ln x \) is employed.

• This leads to \( du = \frac{1}{x} \, dx\), transforming the integral from \( \int \frac{1}{x (\ln x)^p} \, dx\) to \( \int \frac{1}{u^p} \, du\).
• The transformation simplifies the challenge to addressing a straightforward power rule of integration, assisting in determining the convergence behavior of the series.

This clever switch not only reduces the complexity of the integration task but critically aids in revealing the relevant convergence criteria by observing the behavior of the substituted variable as it approaches infinity. Such simplifications are instrumental in tackling advanced convergence problems efficiently.