Problem 47
Question
Show that if \(a_{n}>0\) and \(\lim _{n \rightarrow \infty} n a_{n} \neq 0,\) then \(\Sigma a_{n}\) is divergent.
Step-by-Step Solution
Verified Answer
The series \( \Sigma a_n \) diverges because \( n a_n \neq 0 \) implies \( a_n > \frac{k}{n} \) for some \( k > 0 \), leading to a comparison with the divergent harmonic series.
1Step 1: Understand the Concept of Series and Convergence
A series \( \Sigma a_n \) converges if the sequence of its partial sums \( S_n = a_1 + a_2 + \ldots + a_n \) converges to a finite limit as \( n \to \infty \). If it doesn't, the series diverges.
2Step 2: Examine the Given Condition
We are given \( a_n > 0 \) and \( \lim_{n \to \infty} n a_n eq 0 \). The condition \( \lim_{n \to \infty} n a_n eq 0 \) means that \( n a_n \) does not approach zero; instead, it approaches some nonzero constant or diverges to infinity.
3Step 3: Apply the Divergence Test
For a series \( \Sigma a_n \), if \( \lim_{n \to \infty} a_n eq 0 \), the series diverges. Since \( n a_n \) does not approach zero, \( a_n \) must be greater than \( \frac{k}{n} \) for some \( k > 0 \) and sufficiently large \( n \).
4Step 4: Demonstrate Divergence Using Harmonic Series Comparison
Recall that \( \Sigma \frac{1}{n} \) is a harmonic series and it diverges. Since \( a_n > \frac{k}{n} \) for large \( n \), we have \( \Sigma a_n > k \Sigma \frac{1}{n} \), which implies \( \Sigma a_n \) diverges as it is greater than a divergent series.
Key Concepts
Convergent SeriesDivergence TestHarmonic Series
Convergent Series
When discussing series in mathematics, a major goal is to determine if a series converges or diverges. A series is a sum of sequence of terms. For a series \( \Sigma a_n \) to be considered convergent, the sequence of its partial sums \( S_n = a_1 + a_2 + \ldots + a_n \) must converge to a finite number as \( n \) tends to infinity. Essentially, this means that as you keep adding more terms from the sequence, the total gets closer and closer to a specific value.
If the partial sums do not approach any finite number, the series is divergent. Convergence is important because it means the series has a definitive value, which is particularly useful in various fields like physics and engineering where precise calculations are necessary. When determining convergence, one often employs various tests, such as the ratio test or comparison test, depending on the nature of the series.
If the partial sums do not approach any finite number, the series is divergent. Convergence is important because it means the series has a definitive value, which is particularly useful in various fields like physics and engineering where precise calculations are necessary. When determining convergence, one often employs various tests, such as the ratio test or comparison test, depending on the nature of the series.
Divergence Test
The divergence test is a straightforward tool used to identify whether a series \( \Sigma a_n \) diverges. If the limit \( \lim_{n \to \infty} a_n eq 0 \), then the series must diverge. Simply put, if your sequence terms do not tend towards zero, adding them infinitely will only cause the sum to grow without bound. However, it is crucial to remember that the divergence test cannot prove convergence – it can only confirm divergence.
This test is particularly effective because it provides a quick way to rule out certain series from further examination. In the problem, because \( \lim_{n \to \infty} n a_n eq 0 \), it gives us insight that \( a_n \) does not approach zero, supporting the argument for divergence using further reasoning such as comparisons to known divergent series.
This test is particularly effective because it provides a quick way to rule out certain series from further examination. In the problem, because \( \lim_{n \to \infty} n a_n eq 0 \), it gives us insight that \( a_n \) does not approach zero, supporting the argument for divergence using further reasoning such as comparisons to known divergent series.
Harmonic Series
The harmonic series is a classic example in the study of divergent series. It is represented as \( \Sigma \frac{1}{n} \), involving the sum of reciprocals of natural numbers: \( \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \ldots \). Despite the terms \( \frac{1}{n} \) approaching zero as \( n \) increases, the harmonic series itself diverges. This illustrates a critical point that a series with terms approaching zero isn't enough to ensure convergence.
The problem at hand uses the harmonic series as a comparative tool. Since the given sequence \( a_n \) is greater than \( \frac{k}{n} \) for some constant \( k > 0 \), and the harmonic series diverges, the series \( \Sigma a_n \) diverges as well. This reasoning solidifies the comparison test method where the terms of a series are shown to be larger than those of a known divergent series.
The problem at hand uses the harmonic series as a comparative tool. Since the given sequence \( a_n \) is greater than \( \frac{k}{n} \) for some constant \( k > 0 \), and the harmonic series diverges, the series \( \Sigma a_n \) diverges as well. This reasoning solidifies the comparison test method where the terms of a series are shown to be larger than those of a known divergent series.
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