Question

Does \(\sum_{n=1}^{\infty} 2^{-\ln \ln n}\) converge? (Hint: Write \(2^{\ln \ln n}\) as a power of \(\left.\ln n .\right)\)

Short answer

The series diverges because the terms are larger than those of a convergent p-series with \( p > 1 \).

Step-by-step solution

Analyze the Series

The given series is \( \sum_{n=1}^{\infty} 2^{-\ln \ln n} \). Our task is to determine the convergence of this series.

Simplify the Exponent

Rewrite the exponent using the hint. We have \( 2^{-\ln \ln n} = \left(\frac{1}{2}\right)^{\ln \ln n} = \left(\ln n\right)^{-\ln 2} \) because \( 2^{x} = e^{x \cdot \ln 2} \). So, we can view the terms as \( \left(\ln n\right)^{-\ln 2} \).

Analyze Using Integral Test Mechanism

For large \( n \), \( \ln n \) increases without bound. Consider applying a comparison or integral test. The terms \( \left(\ln n\right)^{-\ln 2} \) are positive and decreasing for large \( n \) as \( \ln 2 > 0 \).

Comparison to a p-Series

Recall that a p-series \( \sum_{n=1}^{\infty} \frac{1}{n^p} \) converges when \( p > 1 \). Since our expression \( \left(\ln n\right)^{-\ln 2} \) can be compared to a series \( \sum \frac{1}{(\ln n)^{p}} \) with \( p = \ln 2 \), we need to confirm if \( \ln 2 > 1 \) or not. Since \( \ln 2 \approx 0.693 < 1 \), the series diverges.

Key concepts

Integral Test and Its Application to Convergence

In the realm of infinite series, determining whether a series converges or diverges can often seem like unraveling a complex puzzle. One of the powerful tools at our disposal is the integral test. This test helps us understand the behavior of series that comprise continuous, positive, and decreasing functions. By comparing a series to an integral, we can draw conclusions about its convergence.

The integral test states that given a function, say \( f(n) \), that is positive, continuous, and decreasing on an interval \([N, \infty)\) for \( N \geq 1 \), if the integral \( \int_N^{\infty} f(x) \, dx \) is convergent, then the series \( \sum_{n=N}^{\infty} f(n) \) is also convergent. Conversely, if the integral diverges, then so does the series.

Applying the integral test to the series \( \sum_{n=1}^\infty (\ln n)^{-\ln 2} \), we leverage the hint that the functions involved are indeed positive and decreasing for large \( n \). However, because the corresponding integral diverges, we confirm that the series \( \sum_{n=1}^{\infty} (\ln n)^{-\ln 2} \) is also divergent. This strategic test provides valuable insights when direct observation falls short.

Understanding Convergence of p-Series

A foundational concept to grasp when dealing with infinite series is that of a p-series. A p-series takes the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \) and is a keystone in analyzing more complex sequences. The key property of such series is that based on the value of \( p \), we can determine convergence or divergence.

A p-series converges if \( p > 1 \) and diverges if \( p \leq 1 \). This rule is derived from comparisons with integrals and is consistent with our understanding of harmonic series and geometric series behavior.

In our given series problem, we transformed the terms to resemble a p-series as \( \sum \frac{1}{(\ln n)^{p}} \), specifically with \( p = \ln 2 \). Since \( \ln 2 \approx 0.693 < 1 \), the comparison clarifies that the series diverges, as it fits the criteria of a p-series with \( p \leq 1 \). Recognizing this comparison is crucial to drawing accurate conclusions in series convergence.

Role of Exponential Functions in Series Analysis

Exponential functions play a vital role in the analysis of series, particularly when it comes to transforming and simplifying complex expressions. Such functions, characterized by constants raised to the power of variables, frequently appear in exponential growth and decay problems.

For the series \(\sum_{n=1}^{\infty} 2^{- ext{ln ln} \, n}\), we utilized the property that \( 2^{x} = e^{x \cdot \ln 2} \) to transform the expression into one more amenable to analysis. By understanding that \( 2^{- ext{ln ln} \, n} = (\ln n)^{- ext{ln} 2} \), we were able to align the terms with a p-series form. This transformation was critical for applying the comparison test, revealing the series' diverging nature.

These manipulations highlight the versatility of exponential functions, enabling us to explore complex sequences by converting them into forms that align with known convergence tests. Mastery of exponential transformations allows seamless entry into various convergence discussions, enriching our mathematical toolbox.