Problem 40 - Calculus: Early Transcendentals

Question

(a) Suppose that \( \sum a_n \) and \( \sum b_n \) are series with positive terms and \( \sum b_n \) is convergent. Prove that if \( \displaystyle \lim_{n \to \infty} \frac {a_n}{b_n} = 0 \) then \( \sum a_n \) is also convergent. (b) Use part (a) to show that the series converges. (i) \( \displaystyle \sum_{n = 1}^{\infty} \frac {\ln n }{n^3} \) (ii) \( \displaystyle \sum_{n = 1}^{\infty} \frac { \ln n}{\sqrt n e^n} \)

Step-by-Step Solution

Answer

Both series converge.

Step 1: Understanding Part (a)

Given the series \( \sum a_n \) and \( \sum b_n \) with positive terms and \( \sum b_n \) is convergent. We want to prove that if \( \lim_{n \to \infty} \frac{a_n}{b_n} = 0 \), then \( \sum a_n \) is convergent.

Step 2: Apply the Limit Comparison Test Conditions

According to the Limit Comparison Test, to show that \( \sum a_n \) converges, we need to show \( \lim_{n \to \infty} \frac{a_n}{b_n} = 0 \) together with the convergence of \( \sum b_n \), implies the convergence of \( \sum a_n \).

Step 3: Conclude with the Limit Comparison Test

Since \( \lim_{n \to \infty} \frac{a_n}{b_n} = 0 \), it indicates that beyond a certain index \( N \), \( a_n < C \cdot b_n \) for all \( n > N \) and any constant \( C > 0 \). Given \( \sum b_n \) converges, by comparison, \( \sum a_n \) must also converge.

Step 4: Applying Part (a) to Series (i)

Consider the series \( \sum_{n = 1}^{\infty} \frac{\ln n}{n^3} \). Choose a comparison series \( b_n = \frac{1}{n^2} \), which is known to converge. Compute \( \lim_{n \to \infty} \frac{\ln n / n^3}{1 / n^2} = \lim_{n \to \infty} \frac{\ln n}{n} = 0 \). This proves that \( \sum_{n=1}^{\infty} \frac{\ln n}{n^3} \) converges due to part (a).

Step 5: Applying Part (a) to Series (ii)

Consider the series \( \sum_{n = 1}^{\infty} \frac{\ln n}{\sqrt{n} e^n} \). Choose a comparison series \( b_n = \frac{1}{e^n} \), which is known to converge. Compute \( \lim_{n \to \infty} \frac{\ln n / \sqrt{n} e^n}{1 / e^n} = \lim_{n \to \infty} \frac{\ln n}{\sqrt{n}} = 0 \). Thanks to part (a), the series \( \sum_{n=1}^{\infty} \frac{\ln n}{\sqrt{n} e^n} \) also converges.

Key Concepts

Limit Comparison Test

The Limit Comparison Test is a powerful tool in determining the convergence of series. It comes in handy particularly when dealing with series where direct evaluation of convergence is complex. Here's how it works in simple terms.

The test involves two series, say \( \sum a_n \) and \( \sum b_n \), both with positive terms. If you know that \( \sum b_n \) converges – which means it adds up to a finite number – you can use the limit comparison test to check whether \( \sum a_n \) does the same.

The core idea is to compute the limit \( \lim_{n \to \infty} \frac{a_n}{b_n} \).

• If this limit equals zero, then it suggests that the terms of \( a_n \) are negligible compared to \( b_n \) for large \( n \), meaning \( \sum a_n \) also converges.
• If the limit is a positive constant (not zero or infinite), then both series \( \sum a_n \) and \( \sum b_n \) converge or diverge together.

The beauty of the Limit Comparison Test lies in its ability to simplify complex functions into a form where convergence is already determined.

Convergent series

A series is said to be convergent if the sum of its infinite terms approaches a finite number as more terms are added.

In practical terms, imagine adding fractions which get smaller and smaller – if those fractions shrink quickly enough to become almost nothing, their sum won't spiral into infinity but eventually stabilize at a certain point. This stabilization is what we call convergence.

There are some well-known series that are convergent, such as the geometric series \( \sum_{n=0}^{\infty} r^n \) when \( |r| < 1 \), and the \( p \)-series \( \sum_{n=1}^{\infty} \frac{1}{n^p} \) with \( p > 1 \).

Knowing which series converge is crucial because other lesser-known or more complicated series can be compared to them, using tests like the Limit Comparison Test to determine their convergence status.

Positive term series

In calculus, a positive term series means each term added is greater than zero.

This is important because series with positive terms have unique convergence behaviors compared to series that may include negative terms.

For example, a positive term series will either converge to a finite sum or diverge to infinity.

• The lack of negative terms makes the application of comparison tests straightforward as there is no concern for terms canceling each other out.

In solving series convergence problems, especially with positive terms, one key step is to often find another series that we already know converges or diverges and use it as a benchmark or reference using tests like the Limit Comparison Test.

This direct comparison offers a clearer path to determining the nature of the series in question without the complications that arise with alternating series.