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Ch. 10 - Infinite Sequences and Series
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 10, Problem 10.4.62

Suppose that aₙ > 0 and limₙ→∞ n²aₙ = 0. Prove that ∑aₙ converges.

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Start by analyzing the given condition: \( \lim_{n \to \infty} n^{2} a_{n} = 0 \). This means that for large \( n \), the term \( a_{n} \) behaves like something smaller than or on the order of \( \frac{1}{n^{2}} \).
Recall the comparison test for series convergence: if \( 0 < a_{n} \leq b_{n} \) for all sufficiently large \( n \), and \( \sum b_{n} \) converges, then \( \sum a_{n} \) also converges.
Since \( \lim_{n \to \infty} n^{2} a_{n} = 0 \), for any \( \varepsilon > 0 \), there exists \( N \) such that for all \( n > N \), \( n^{2} a_{n} < \varepsilon \). Choose \( \varepsilon = 1 \) to get \( a_{n} < \frac{1}{n^{2}} \) for all \( n > N \).
We know that the series \( \sum_{n=1}^{\infty} \frac{1}{n^{2}} \) converges (it is a p-series with \( p = 2 > 1 \)). Therefore, by the comparison test, \( \sum_{n=1}^{\infty} a_{n} \) converges.
Conclude that since \( a_{n} > 0 \) and \( a_{n} \) is eventually bounded above by a convergent p-series term, the series \( \sum a_{n} \) converges.

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Key Concepts

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

Limit of a Sequence

The limit of a sequence describes the value that the terms of the sequence approach as the index goes to infinity. In this problem, understanding that limₙ→∞ n²aₙ = 0 means the terms aₙ shrink faster than 1/n² is crucial for comparing series.
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Introduction to Sequences

Comparison Test for Series Convergence

The comparison test states that if 0 ≤ aₙ ≤ bₙ for all n beyond some index and ∑bₙ converges, then ∑aₙ also converges. This test helps prove convergence by comparing aₙ to a known convergent series like 1/n².
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Direct Comparison Test

p-Series and Its Convergence

A p-series ∑1/n^p converges if and only if p > 1. Since the problem involves terms related to 1/n², recognizing that ∑1/n² converges provides a benchmark to show that ∑aₙ converges by comparison.
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P-Series and Harmonic Series