Problem 51 - Thomas Calculus

Question

Recursively Defined Terms Which of the series \(\sum_{n=1}^{\infty} a_{n}\) defined by the formulas in Exercises \(45-54\) converge, and which diverge? Give reasons for your answers. $$a_{1}=1, \quad a_{n+1}=\frac{1+\ln n}{n} a_{n}$$

Step-by-Step Solution

Answer

The series diverges as terms approach zero slower than the harmonic series.

Step 1: Understanding the Sequence

The sequence is defined recursively with a starting term \(a_1 = 1\) and the recursive formula \(a_{n+1} = \frac{1 + \ln n}{n} a_n\). We need to assess whether the series \(\sum_{n=1}^{\infty}a_n\) converges or diverges.

Step 2: Analyze Recursive Formula

Examine the recursive formula \(a_{n+1} = \frac{1+\ln n}{n} a_n\). For large \(n\), \(\ln n\) grows slowly, so \(\frac{1 + \ln n}{n}\) generally behaves like \(\frac{\ln n}{n}\), which tends to zero as \(n\) increases.

Step 3: Approximation of the Terms

Recognize that if \(b_n = \frac{1}{n}\), then the recursion becomes approximately \(a_{n+1} \approx \frac{n b_n}{n}a_n = b_n a_n\). Thus, we look at how the terms behave: \(a_{n+1} \approx \frac{\ln n}{n} a_n \approx \frac{a_n}{n}\).

Step 4: Apply Ratio Test for Divergence

Use the Ratio Test. Consider the limit \(L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{1+\ln n}{n} \right|\). Since \(\lim_{n \to \infty} \frac{1}{n} = 0\) and \(\ln n\) grows slower than \(n\), the term \(\frac{1+\ln n}{n}\) approaches zero, indicating the terms decrease to zero.

Step 5: Convergence Assessment

Since the \(\frac{1+\ln n}{n}\) term becomes smaller than any constant, the sequence terms decrease but not necessarily fast enough for the series to converge by the Ratio Test. We further analyze or speculate convergence behavior by comparing it to a known divergent series.

Step 6: Compare with Harmonic Series

A convenient comparison is to compare the convergence of the terms \(a_n\) with the harmonic series \(\sum_{n=1}^{\infty} \frac{1}{n}\), which is divergent. Since \(a_n\) does not approach zero faster than \(\frac{1}{n}\), the series \(\sum_{n=1}^{\infty}a_n\) is divergent.

Key Concepts

Convergence and Divergence

When dealing with series, determining whether they converge or diverge is crucial. Convergence means that as you add more terms, the series approaches a specific limit. Divergence means the series grows indefinitely or does not approach any limit. For example, a series like \(\sum_{n=1}^{\infty} a_n\) converges if the sum reaches a finite number as \(n\) increases. Conversely, it diverges if no such limit exists.

In the given problem, understanding convergence or divergence helps us decide if the series \(\sum_{n=1}^{\infty} a_n\) reaches a finite sum. We look at each term's behavior as \(n\) becomes large, examining if they get small enough fast enough. If the terms don't shrink rapidly, like getting smaller than some diminishing constant, the series is more likely to diverge. Thus, finding the overall behavior of the terms in a series allows us to predict whether they converge or diverge.

Ratio Test

The Ratio Test is a powerful tool for analyzing convergence of series. It uses the concept of looking at the ratio between consecutive terms in a series. Consider the limit \(L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\). This ratio reveals how fast the terms shrink.

- If \(L < 1\), the series converges. The terms are getting smaller quickly.
- If \(L > 1\) or \(L = \infty\), the series diverges. The terms don't decrease sufficiently; they might even grow.
- If \(L = 1\), the test is inconclusive. Further analysis is needed.

In our original exercise, applying the Ratio Test means evaluating \(L = \lim_{n \to \infty} \left| \frac{1 + \ln n}{n} \right|\). As \(n\) grows, the ratio approaches \( \frac{\ln n}{n} \), which tends to zero, but it doesn't guarantee convergence. The terms may become small, but not fast enough to ensure the series converges. We discover that further comparison with other known series is necessary to reach a conclusion.

Harmonic Series Comparison

Comparing a series with another already known series, such as the harmonic series, helps in understanding its convergence behavior. The harmonic series is an example of a divergent series. It takes the form \(\sum_{n=1}^{\infty} \frac{1}{n}\), which means adding terms like \(\frac{1}{1} + \frac{1}{2} + \frac{1}{3}\) and so on. Despite each term being smaller than before, the harmonic series doesn't converge to a specific value; it grows infinitely.

In the original problem, comparing \(a_n\) with the harmonic series gives insights into divergence. If the terms \(a_n\) shrink slower than or similar to \(\frac{1}{n}\), the series \(\sum_{n=1}^{\infty} a_n\) cannot converge. Hence, by evaluating \(a_n\) against \(\frac{1}{n}\), we notice that \(\frac{1+\ln n}{n}\) doesn't decrease sufficiently faster than \(\frac{1}{n}\). This comparison strengthens our conclusion that the series diverges by showing \(a_n\)'s behavior relative to the divergent harmonic series. Aiding to understand why some series just won't add up to a limit.