Cauchy condensation
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![In this homework, we'll learn about a new convergence test called **Cauchy's Condensation Test**. Suppose \(\sum_{n=1}^{\infty} a_n\) is a series where the terms \(a_n\) are \(\geq 0\) and decreasing, so \(a_{n+1} \leq a_n\) for all \(n\). The Cauchy condensation test says that
\[
\sum_{n=1}^{\infty} a_n \text{ converges if and only if } \sum_{n=1}^{\infty} 2^n a_{2^n} \text{ converges.}
\]
This test generalizes the idea from Oresme's proof of the divergence of the harmonic series where you group terms together in groups of size 1, 2, 4, 8, etc.
1. By grouping the terms \(a_n\) together into groups of size 1, 2, 4, \dots, \(2^N\), argue that
\[
\sum_{n=1}^{2^{N+1}-1} a_n \leq \sum_{n=0}^{N} 2^n a_{2^n}.
\]
2. By taking the limit \(N \to \infty\), argue that if \(\sum_{n=1}^{\infty} 2^n a_{2^n}\) converges, then so does \(\sum_{n=1}^{\infty} a_n\). (Hint: apply the monotone convergence theorem to the sequence of partial sums of the series \(\sum_{n=1}^{\infty} a_n\).)
This is half of the proof that the Cauchy condensation test works. A similar argument can be used to show that if \(\sum_{n=1}^{\infty} 2^n a_{2^n}\) diverges, then so does \(\sum_{n=1}^{\infty} a_n\). You don't have to show this.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5a238e39-c297-4642-8cf7-f442113dee2f%2Fa215ea21-6d1c-4867-9d93-e26e628b8519%2Fw64bu3n_processed.jpeg&w=3840&q=75)
Transcribed Image Text:In this homework, we'll learn about a new convergence test called **Cauchy's Condensation Test**. Suppose \(\sum_{n=1}^{\infty} a_n\) is a series where the terms \(a_n\) are \(\geq 0\) and decreasing, so \(a_{n+1} \leq a_n\) for all \(n\). The Cauchy condensation test says that
\[
\sum_{n=1}^{\infty} a_n \text{ converges if and only if } \sum_{n=1}^{\infty} 2^n a_{2^n} \text{ converges.}
\]
This test generalizes the idea from Oresme's proof of the divergence of the harmonic series where you group terms together in groups of size 1, 2, 4, 8, etc.
1. By grouping the terms \(a_n\) together into groups of size 1, 2, 4, \dots, \(2^N\), argue that
\[
\sum_{n=1}^{2^{N+1}-1} a_n \leq \sum_{n=0}^{N} 2^n a_{2^n}.
\]
2. By taking the limit \(N \to \infty\), argue that if \(\sum_{n=1}^{\infty} 2^n a_{2^n}\) converges, then so does \(\sum_{n=1}^{\infty} a_n\). (Hint: apply the monotone convergence theorem to the sequence of partial sums of the series \(\sum_{n=1}^{\infty} a_n\).)
This is half of the proof that the Cauchy condensation test works. A similar argument can be used to show that if \(\sum_{n=1}^{\infty} 2^n a_{2^n}\) diverges, then so does \(\sum_{n=1}^{\infty} a_n\). You don't have to show this.
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