Problem 42 - Essential Calculus Early Transcendentals

Question

(a) Suppose that \(\Sigma a_{n}\) and \(\Sigma b_{n}\) are series with positive terms and \(\Sigma b_{n}\) is convergent. Prove that if \(\lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}=0\) then \(\Sigma a_{n}\) is also convergent. (b) Use part (a) to show that the series converges. $$\sum_{n=1}^{\infty} \frac{\ln n}{n^{3}} \quad \text { (ii) } \sum_{n=1}^{\infty} \frac{\ln n}{\sqrt{n} e^{n}} $$

Step-by-Step Solution

Answer

Both series \(\sum_{n=1}^{\infty} \frac{\ln n}{n^{3}}\) and \(\sum_{n=1}^{\infty} \frac{\ln n}{\sqrt{n} e^{n}}\) converge by the Limit Comparison Test.

Step 1: Understanding Given Conditions

We begin by analyzing the provided conditions: Two series \(\Sigma a_{n}\) and \(\Sigma b_{n}\) have positive terms, and the series \(\Sigma b_{n}\) is convergent. Also, it is given that \(\lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}=0\). Our objective is to prove that \(\Sigma a_{n}\) is convergent under these conditions.

Step 2: Applying the Limit Comparison

To use the limit \(\lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}=0\), let's think in terms of the limit comparison test. The limit condition suggests that for sufficiently large \(n\), \(a_n\) is much smaller than \(b_n\). Since \(\Sigma b_{n}\) converges, these insights imply \(\Sigma a_{n}\) must also converge.

Step 3: Formalizing the Argument

For every \(\varepsilon > 0\), there exists a number \(N\) such that for all \(n > N\), \(\left|\frac{a_{n}}{b_{n}}\right| < \varepsilon\). Set \(\varepsilon = 0.5\), which implies \(a_{n} < 0.5 b_{n}\), for all \(n > N\). Because \(\Sigma b_{n}\) converges, by the comparison test, since \(0 < a_n < 0.5 b_n\) for \(n > N\), \(\Sigma a_{n}\) must also converge.

Step 4: Application on (i) Series

Consider the series \(\sum_{n=1}^{\infty} \frac{\ln n}{n^{3}}\). Set \(a_{n} = \frac{\ln n}{n^{3}}\) and let \(b_{n} = \frac{1}{n^{2.5}}\). Note \(\Sigma b_{n}\) converges by the p-series test (p = 2.5 > 1). Show \(\lim_{n \to \infty} \frac{a_n}{b_n} = 0\) using L'Hôpital's Rule or any suitable simplification.

Step 5: Application on (ii) Series

Now consider \(\sum_{n=1}^{\infty} \frac{\ln n}{\sqrt{n} e^{n}}\). Let \(a_n = \frac{\ln n}{\sqrt{n} e^{n}}\) and \(b_n = \frac{1}{e^{n/2}}\). A series with \(b_n = \frac{1}{e^{n/2}}\) will converge because it behaves like a geometric series with common ratio \(e^{-1/2}\). Demonstrate \(\lim_{n \to \infty} \frac{a_n}{b_n} = 0\), showing \(\Sigma a_{n}\) converges.

Key Concepts

Limit Comparison Test

The Limit Comparison Test is a helpful tool for determining the convergence of a series. It relies on comparing two series with positive terms: \(\Sigma a_n\) and \(\Sigma b_n\). If \(\lim_{n \to \infty} \frac{a_n}{b_n} = L\), where \(0 < L < \infty\), both series either converge or diverge together. However, if \(\lim_{n \to \infty} \frac{a_n}{b_n} = 0\) and \(\Sigma b_n\) is known to converge, \(\Sigma a_n\) must also converge. This test helps when direct analysis is tricky, by leveraging another series that is easier to understand.For example, in the given problem, we are told that \(\lim_{n \to \infty} \frac{a_n}{b_n} = 0\). This means, eventually, \(a_n\) becomes much smaller than \(b_n\). Since \(\Sigma b_n\) converges, \(\Sigma a_n\) must also converge based on the premise of the Limit Comparison Test.

p-Series Test

The p-Series Test is vital for identifying the convergence of series in the form \(\sum_{n=1}^{\infty} \frac{1}{n^p}\). According to this test:

• If \(p > 1\), the series converges.
• If \(p \leq 1\), the series diverges.

Consider \(\sum_{n=1}^{\infty} \frac{1}{n^{2.5}}\), which is a p-series with \(p = 2.5\). Because 2.5 is greater than 1, we conclude this series converges. This principle is used to determine the behavior of more complex series when comparing them using the Limit Comparison Test. In the given exercise, identifying a related p-series and knowing its convergence helps validate other series by comparison.

Geometric Series

Geometric series is a cornerstone in understanding series convergence. A geometric series is of the form \(\sum_{n=0}^{\infty} ar^n\), where \(a\) is the first term and \(r\) is the common ratio. The convergence criteria for geometric series are:

• It converges if \(|r| < 1\).
• It diverges if \(|r| \geq 1\).

In part (ii) of the exercise, the series component \(\frac{1}{e^{n/2}}\) acts like a geometric series, whereby each subsequent term is multiplied by a constant ratio, specifically \(e^{-1/2}\). Given the ratio \(|e^{-1/2}| < 1\), the series converges. This insight about geometric series being used demonstrates its utility in evaluating more complex expressions.

L'Hôpital's Rule

L'Hôpital's Rule is instrumental when determining limits, particularly when faced with indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). The rule states that:If \(\lim_{x \to a} f(x) = 0\) and \(\lim_{x \to a} g(x) = 0\), or both are infinite, \[\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}\](if the limit on the right side exists).This rule is frequently applied to simplify the limit expressions encountered in the Limit Comparison Test. In step (iv) of the solution process, L'Hôpital's Rule assists in showing that the limit \(\lim_{n \to \infty} \left(\frac{\ln n}{n^3}\right) / \left(\frac{1}{n^{2.5}}\right)\) approaches zero, confirming that \(\Sigma a_n\) converges as \(\Sigma b_n\) is a known convergent series. This powerful tool breaks down complex limit problems to more straightforward derivative expressions.