Problem 35 - Calculus: Early Transcendentals

Question

Test the series for convergence or divergence. \( \displaystyle \sum_{n = 1}^{\infty} \frac {1}{n^{1+1/n}} \)

Step-by-Step Solution

Answer

The series diverges.

Step 1: Identify the Series Type

The given series is \( \sum_{n=1}^{\infty} \frac{1}{n^{1+1/n}} \). This looks similar to a harmonic series, which has the form \( \sum_{n=1}^{\infty} \frac{1}{n} \), but with a different exponent in the denominator.

Step 2: Analyze the Exponent in the Denominator

The exponent in the denominator is \( 1 + \frac{1}{n} \). For large \( n \), \( \frac{1}{n} \) approaches 0, making the exponent approach 1. This analysis suggests that the terms of the series \( \frac{1}{n^{1+1/n}} \) behave approximately like \( \frac{1}{n} \), but with minor adjustments.

Step 3: Consider Using the Limit Comparison Test

Since the terms \( \frac{1}{n^{1+1/n}} \) behave similarly to \( \frac{1}{n} \), we can use the Limit Comparison Test with the harmonic series \( \sum \frac{1}{n} \), which is known to diverge.

Step 4: Apply the Limit Comparison Test

Compute the limit \( \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{1}{n^{1+1/n}}}{\frac{1}{n}} = \lim_{n \to \infty} n^{1/n} \). Note that \( n^{1/n} = e^{\ln(n)/n} \), and as \( n \to \infty \), \( \ln(n)/n \to 0 \), making \( e^{\ln(n)/n} \to e^0 = 1 \).

Step 5: Conclude from the Limit Comparison Test

Since the limit is 1, which is a positive finite number, the Limit Comparison Test tells us that \( \sum \frac{1}{n^{1+1/n}} \) and \( \sum \frac{1}{n} \) behave similarly. Since \( \sum \frac{1}{n} \) diverges, \( \sum \frac{1}{n^{1+1/n}} \) also diverges.

Key Concepts

Limit Comparison Test

The Limit Comparison Test is a valuable tool in determining the convergence or divergence of series. When you have two series, say \( a_n \) and \( b_n \), you'd want to see how they compare to one another as \( n \to \infty \). Here's the key idea:

• If \( \lim_{n \to \infty} \frac{a_n}{b_n} = c \), where \( c \) is a positive finite number, both series \( \sum a_n \) and \( \sum b_n \) either converge or diverge together.

In our particular case, the series \( a_n = \frac{1}{n^{1+1/n}} \) was compared to the well-known harmonic series \( b_n = \frac{1}{n} \). The limit turned out to be 1, a positive finite number, so the behavior of \( \sum a_n \) mirrors that of \( \sum b_n \). Since the harmonic series diverges, so does \( \sum \frac{1}{n^{1+1/n}} \). This makes the Limit Comparison Test both powerful and straightforward for analyzing series with terms that resemble known forms.

Harmonic Series

The harmonic series is one of the classic divergent series in mathematics. It is expressed as:

• \( \sum_{n=1}^{\infty} \frac{1}{n} \)

Despite each term \( \frac{1}{n} \) getting smaller and smaller, the series does not converge to a finite number. Each additional term contributes a small value, but these contributions accumulate to infinity. The divergence of the harmonic series is due to its very slow rate of decay. Understanding this helps when comparing similar series, like the one in question \( \sum \frac{1}{n^{1+1/n}} \), where the behavior is close enough to the harmonic series to infer divergence.

Divergence in Series

Divergence in series is an important concept in mathematical analysis. Simply put, a series diverges if the sum of its terms doesn't settle down to a fixed, finite limit as more terms are added. Instead, it either increases indefinitely or oscillates without approaching a specific value.

• The fact that a series like \( \sum \frac{1}{n} \) diverges despite terms getting smaller demonstrates the importance of understanding the nature of terms, rather than just their size.
• The key takeaway is that even infinitely small terms can sum to infinity, highlighting why convergence tests are essential.

Applying divergence concepts, as seen in the series \( \sum \frac{1}{n^{1+1/n}} \), helps in determining behavior based on term analysis and series comparisons.

Asymptotic Behavior

In mathematics, asymptotic behavior describes how functions behave as inputs get very large. This is especially relevant in series analysis where the behavior of terms for large \( n \) can determine convergence or divergence.

• In our series, \( \frac{1}{n^{1+1/n}} \) behaves asymptotically like \( \frac{1}{n} \), because for large \( n \), the term \( \frac{1}{n} \) in the exponent becomes negligible.

Understanding this helps in predicting series behavior without necessarily calculating infinite sums explicitly. Seeing terms shrink at rates equivalent to known divergent forms cues us into applying appropriate tests.

Exponential Limits

Exponential limits are useful in analyzing series, especially when terms involve exponential functions. The essential principle arises when evaluating the asymptotic expression of a function:

• For \( n^{1/n} \), we use the transformation \( e^{\ln(n)/n} \), which simplifies analysis as \( \ln(n)/n \to 0 \) for large \( n \).

This transformation shows that \( n^{1/n} \) approaches 1 as \( n \to \infty \), making the limit of \( \frac{a_n}{b_n} \) easy to evaluate. It's a neat trick that transforms complicated-looking terms into manageable limits, aiding in Limit Comparison Test applications, as demonstrated in the original series problem.