Skip to main content
Ch. 10 - Infinite Sequences and Series
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 10, Problem 10.3.62a

(Continuation of Exercise 61.) Use the result in Exercise 61 to determine which of the following series converge and which diverge. Support your answer in each case.
a. ∑ (from n=2 to ∞) [1 / (n ln n)]

Verified step by step guidance
1
Recall the result from Exercise 61, which likely involved the Integral Test for series convergence, especially for series of the form \( \sum \frac{1}{n (\ln n)^p} \). The Integral Test states that if \( f(x) = \frac{1}{x \ln x} \) is positive, continuous, and decreasing for \( x \geq 2 \), then the convergence of the series \( \sum_{n=2}^\infty \frac{1}{n \ln n} \) is determined by the convergence of the integral \( \int_2^\infty \frac{1}{x \ln x} \, dx \).
Set up the integral to apply the Integral Test: \[ \int_2^\infty \frac{1}{x \ln x} \, dx. \] This integral will help us determine if the series converges or diverges.
Use the substitution method to evaluate the integral: let \( u = \ln x \), so that \( du = \frac{1}{x} dx \). This transforms the integral into \[ \int_{\ln 2}^\infty \frac{1}{u} \, du. \]
Recognize that \( \int \frac{1}{u} \, du = \ln |u| + C \). Therefore, the integral becomes \[ \lim_{t \to \infty} \int_{\ln 2}^t \frac{1}{u} \, du = \lim_{t \to \infty} (\ln t - \ln (\ln 2)) = \infty. \]
Since the integral diverges to infinity, by the Integral Test, the series \( \sum_{n=2}^\infty \frac{1}{n \ln n} \) also diverges.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
4m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Comparison Test for Series

The Comparison Test helps determine convergence or divergence by comparing a given series to another series with known behavior. If the terms of the given series are smaller than those of a convergent series, it also converges; if larger than those of a divergent series, it diverges.
Recommended video:
09:25
Direct Comparison Test

Integral Test

The Integral Test relates the convergence of a series to the convergence of an improper integral of a related function. If the integral of f(x) from some point to infinity converges, then the series ∑ f(n) converges; if the integral diverges, so does the series.
Recommended video:
07:25
Integral Test

Behavior of the Harmonic Series and Logarithmic Modifications

The harmonic series ∑ 1/n diverges, but adding logarithmic terms in the denominator, such as ∑ 1/(n ln n), affects convergence. Understanding how slowly the logarithm grows is key to analyzing whether such series converge or diverge.
Recommended video:
04:30
P-Series and Harmonic Series