Problem 34 - Thomas Calculus

Question

Which of the series in Exercises \(11-40\) converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series' convergence or divergence.) $$ \sum_{n=1}^{\infty} n \tan \frac{1}{n} $$

Step-by-Step Solution

Answer

The series diverges because its terms do not approach zero.

Step 1: Analyze the terms of the series

The series given is \( \sum_{n=1}^{\infty} n \tan \frac{1}{n} \). As \( n \to \infty \), we need to examine the behavior of the terms \( n \tan \frac{1}{n} \) to determine if they approach zero, which is a necessary condition for convergence.

Step 2: Use a Taylor Expansion Approximation

To understand the behavior of \( \tan \frac{1}{n} \) for large \( n \), use the Taylor expansion: \( \tan x \approx x + \frac{x^3}{3} + \text{higher order terms} \text{ for small } x \). Applying this for \( x = \frac{1}{n} \), we get \( \tan \frac{1}{n} \approx \frac{1}{n} + \frac{1}{3n^3} + \ldots \).

Step 3: Substitute the Approximation

Substitute the approximation back into the series: \( n \tan \frac{1}{n} \approx n \left( \frac{1}{n} + \frac{1}{3n^3} \right) = 1 + \frac{1}{3n^2} \). So, for large \( n \), \( n \tan \frac{1}{n} \approx 1 \).

Step 4: Apply the Divergence Test

For a series \( \sum a_n \) to converge, it is necessary that \( \lim_{n \to \infty} a_n = 0 \). We have found that \( \lim_{n \to \infty} n \tan \frac{1}{n} = 1 eq 0 \).

Step 5: Conclude the Result

Since the terms of the series do not approach zero, the series does not meet the necessary condition for convergence. Therefore, by the Divergence Test, the series \( \sum_{n=1}^{\infty} n \tan \frac{1}{n} \) diverges.

Key Concepts

Divergence Test

The Divergence Test is a simple tool to determine whether a series converges or diverges. As a first step when examining a series \( \sum a_n \), check if the terms \( a_n \) approach zero as \( n \) approaches infinity. If \( \lim_{n \to \infty} a_n eq 0 \), the series definitely diverges. This test is essential because, regardless of any other factors, a series cannot converge unless the terms approach zero.

In the case of our example, \( \sum_{n=1}^{\infty} n \tan \frac{1}{n} \), we found that the limit of the terms does not equal zero \( \lim_{n \to \infty} n \tan \frac{1}{n} = 1 \), thereby clearly indicating divergence.

While the Divergence Test can conclusively determine if a series diverges, it does not confirm convergence when \( \lim_{n \to \infty} a_n = 0 \), as there are other characteristics of series that must also be analyzed for convergence.

Taylor Expansion

Taylor Expansion is an invaluable concept in calculus for approximating functions using polynomials. This expansion is particularly powerful for analyzing functions close to a point, often when dealing with small inputs. For a function \( f(x) \) with derivatives of all orders, its Taylor expansion around \( x = 0 \) is given by: \[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \ldots \]

In analyzing the series \( \sum_{n=1}^{\infty} n \tan \frac{1}{n} \), we use the Taylor expansion of the tangent function. For small \( x \), \( \tan x \approx x + \frac{x^3}{3} \). By substituting \( x = \frac{1}{n} \), we estimate \( \tan \frac{1}{n} \approx \frac{1}{n} + \frac{1}{3n^3} \). This approximation allows us to express the term \( n \tan \frac{1}{n} \) as \( 1 + \frac{1}{3n^2} \).

Knowing that higher order terms become negligible for large \( n \), Taylor expansions enable simplifications that reveal underlying behaviors of functions, making them pivotal in understanding series and sequences.

Limit of a Sequence

The limit of a sequence involves finding the value that the terms of a sequence approach as \( n \) tends to infinity. This concept is crucial for determining whether a sequence converges or diverges, which is necessary for analyzing series.

To say that a sequence \( \{a_n\} \) converges to a limit \( L \), for any small positive number \( \varepsilon \), however tiny, there exists a number \( N \) such that for all \( n > N \), the distance \(|a_n - L| < \varepsilon\). A sequence that fails to meet this criterion diverges.

In our example of \( n \tan \frac{1}{n} \), the terms approach 1 as \( n \to \infty \), showing they don't converge to the necessary 0. This failure indicates that the sequence corresponding to series terms diverges. Consequently, the investigation of limits is foundational in sequence and series analysis for determining convergence or divergence.