Show that ∑k=2^∞ (1)/((lnk)^lnk) and ∑k=2^∞ (1)/((lnk)^k) both converge.

Name: Problem 43, Chapter 8: Calculus: Early Transcendental Functions
Uploaded: 2025-10-04T01:52:31-08:00
Duration: 2 min 44 s
Channel: Nick Johnson
Description: Show that ∑k=2^∞ (1)/((lnk)^lnk) and ∑k=2^∞ (1)/((lnk)^k) both converge.

Show that ∑k=2^∞ (1)/((lnk)^lnk) and ∑k=2^∞ (1)/((lnk)^k)...

Key Concepts

Convergence Tests for Infinite Series

This concept involves determining whether an infinite series converges or diverges by employing criteria such as the comparison test, the ratio test, or the root test. In problems like this, one typically establishes bounds or reformulates series terms to compare them with terms of a known convergent or divergent series.

Exponential and Logarithmic Transformations

Problems with terms containing logarithms raised to logarithmic or other exponents can be greatly simplified by rewriting the expressions in exponential form. For example, by expressing expressions such as (ln k)^(ln k) as an exponential function, one can more clearly analyze the rate at which the terms decrease. This transformation is essential in understanding the rapid decay in the series terms.

Asymptotic Growth and Rate of Decay

Understanding how functions grow is crucial in series convergence problems. In this context, analyzing the growth rates of functions like the natural logarithm compared to exponential functions helps to explain why series terms decrease sufficiently fast. Recognizing that terms such as 1/(ln k)^(ln k) or 1/(ln k)^k decay faster than typical polynomial or exponential terms is key to proving convergence.

Transcript

00:01 Yes, so for the first series, we have the series where we have k going from 2 to infinity of 1 over the natural log of k, raised 2 the natural log of k.

00:14 So we can analyze each term, and then we can say that it's going to decrease much faster than any polynomial term, because for large k, you can write this as the natural log of k times the natural log of k, the natural log of k.

00:30 Is then going to approach infinity.

00:33 And then we can compare to a p series where we notice that the natural log of k to the natural log of k is greater than or equal to the natural log of k squared for all k greater than equal to e squared.

00:46 And since we have the natural log of k, the natural log of k is greater than or equal to two, which is going to imply that the natural log of k then is greater than or equal to two.

00:58 Yeah, so for large k, we have that a sub k, which is going to be equal to 1 over the natural log of k raised to the natural log of k, is less than equal to 1 over the natural log of k squared.

01:16 And then for the convergence of comparison, the series 1 over ln of k squared converges.

01:24 It can be checked with the integral test where we have the integral from 2 to infinity.

01:30 Of, let's say, so dx over the natural log of x squared is going to be less than infinity.

01:38 And then by comparison, the sum also is going to converge.

01:43 So the first is going to converge.

01:45 And then for the second series, we analyze the general term for number two here, b sub k, b sub k, which is equal to 1 over the natural log of k raised to the k...

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