Suppose that aₙ \u003e 0 and limₙ→∞ n²aₙ = 0. Prove that ∑aₙ converges.

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 $$ 0 0 $$, there exists $$ N $$ such that for all $$ n \u003e N $$, $$ n^{2} a_{n} N . We $$know that the series $$ \sum_{n=1}^{\infty} \frac{1}{n^{2}} $$ converges (it is a p-series with $$ p = 2 \u003e 1 ). $$Therefore, by the comparison test, $$ \sum_{n=1}^{\infty} a_{n} $$ converges. Conclude that since $$ a_{n} \u003e 0 $$ and $$ a_{n} is $$eventually bounded above by a convergent p-series term, the series $$ \sum a_{n} $$ converges.

Ch. 10 - Infinite Sequences and Series

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Hass 15th Edition

Ch. 10 - Infinite Sequences and Series

문제 10.4.62

10장, 문제 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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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

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.

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.

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.