Problem 16 - Calculus: Early Transcendentals

Question

Determine whether the series is convergent or divergent. \( \displaystyle \sum_{n = 1}^{\infty} \frac {\sqrt n}{1 + n^{3/2}} \)

Step-by-Step Solution

Answer

The series is divergent.

Step 1: Identify the Series

The given series is \( \sum_{n=1}^{\infty} \frac{\sqrt{n}}{1 + n^{3/2}} \). For convergence or divergence, preliminary tests like the convergence test can be applied.

Step 2: Apply the Limit Comparison Test

To apply the Limit Comparison Test, compare it with a similar, simpler series \( \sum_{n=1}^{\infty} \frac{1}{n} \). The term of this chosen series is similar to the given series in form, especially for large \( n \). The general term becomes dominant like \( \frac{1}{n} \) when \( n \to \infty \).

Step 3: Calculate the Limit

Compute the limit \( \lim_{n \to \infty} \frac{ \frac{\sqrt{n}}{1+n^{3/2}} }{ \frac{1}{n} } = \lim_{n \to \infty} \frac{n^{3/2}}{1+n^{3/2}} \).

Step 4: Simplify the Limit Expression

Simplify \( \lim_{n \to \infty} \frac{n^{3/2}}{1+n^{3/2}} = \lim_{n \to \infty} \frac{1}{\frac{1}{n^{3/2}} + 1} \). As \( n \to \infty \), \( \frac{1}{n^{3/2}} \to 0 \), simplifying to \( \lim_{n \to \infty} \frac{1}{0+1} = 1 \).

Step 5: Evaluate Convergence

Since the limit \( L = 1 \) and \( 0 < L < \infty \), the Limit Comparison Test implies that the original series \( \sum_{n=1}^{\infty} \frac{\sqrt{n}}{1 + n^{3/2}} \) converges or diverges together with the simpler series \( \sum \frac{1}{n} \), which is known to diverge.

Step 6: Conclusion

Since \( \sum \frac{1}{n} \) is divergent, the original series \( \sum \frac{\sqrt{n}}{1 + n^{3/2}} \) is also divergent.

Key Concepts

Limit Comparison Test

The Limit Comparison Test is a powerful tool for determining whether a given series converges or diverges. This test requires you to compare the series in question with another series whose convergence behavior is well known. To utilize this test, you need to formulate the ratio of the terms of the two series and find the limit as the term index approaches infinity.

If this limit, which we'll call \( L \), exists and is positive (i.e., \( 0 < L < \infty \)), then both series will either converge or diverge together. This makes the Limit Comparison Test especially useful when dealing with complex series where direct evaluation is difficult.

For example, in the original problem, the series \( \sum_{n=1}^{\infty} \frac{\sqrt{n}}{1+n^{3/2}} \) was compared with \( \sum_{n=1}^{\infty} \frac{1}{n} \). By performing the limit computation for their ratio and obtaining a result of 1, a known positive constant, it is concluded that both series share the same convergence behavior. Since \( \sum \frac{1}{n} \) diverges, the original series also diverges.

Series Convergence

Convergence in the context of series specifically refers to the terms of the series approaching a single value (a finite limit) as the number of terms increases indefinitely. For a series \( \sum a_n \) to converge, the limit of the sequence of partial sums \( S_n = a_1 + a_2 + a_3 + \cdots + a_n \) must exist and be finite as \( n \to \infty \).

The convergence of a series is a fundamental topic because it allows us to sum infinitely many numbers yet still obtain a precise finite result. Various tests exist to help determine convergence, including the Limit Comparison Test, which is particularly useful for series that are difficult to analyze directly. Other tests include the Integral Test and the Ratio Test, among others.

Understanding how to analyze the convergence of a series allows students to actually "tame the infinite" and have predictive control over how sequences and series behave. This concept is crucial for further studies in calculus, complex analysis, and beyond.

Infinite Series

An infinite series is simply the sum of an infinite sequence of numbers. When we write \( \sum_{n=1}^{\infty} a_n \), it indicates the addition of terms \( a_1, a_2, a_3, \ldots \), extending indefinitely. The study of infinite series is vital in mathematics because it provides us the tools to evaluate sums that are not easily computed in finite steps.

Infinite series can exhibit either convergence or divergence. A series converges if, as you add more terms, the series approaches a specific value. Conversely, if it does not settle towards any limit, it diverges.

Convergent series often have practical applications. For example, they are pivotal in representing functions in power series form, a cornerstone in both pure and applied mathematics. Additionally, divergent series also have mathematical importance, as learning about them ensures we avoid erroneous assumptions about infinite processes.

• In many cases, tests like the Limit Comparison Test helps understand an infinite series' behavior without evaluating every term individually.

Understanding infinite series lays the groundwork for more advanced topics in calculus and other branches of mathematics, providing insight into how infinite processes can be thoroughly analyzed and understood.