Problem 8 - Single Variable Calculus

Question

Determine whether the series converges or diverges. $$ \sum_{n=2}^{\infty} \frac{(-1)^{n-1}}{\ln n} $$

Step-by-Step Solution

Answer

The series \(\sum_{n=2}^{\infty} \frac{(-1)^{n-1}}{\ln n}\) converges because it satisfies both conditions of the Alternating Series Test: \(\lim_{n\to\infty} \frac{1}{\ln n} = 0\) and the sequence \(a_n = \frac{1}{\ln n}\) is decreasing.

Step 1: Identify the alternating series type

First, let's recognize that this is an alternating series. An alternating series has the form: \[ \sum_{n=1}^{\infty} (-1)^{n-1} a_n \text{ or } \sum_{n=1}^{\infty} (-1)^n a_n \] In our case, the series is given by: \[ \sum_{n=2}^{\infty} \frac{(-1)^{n-1}}{\ln n} \]

Step 2: Apply the Alternating Series Test

The Alternating Series Test states that a series of the form \(\sum_{n=1}^{\infty} (-1)^{n-1} a_n\) or \(\sum_{n=1}^{\infty} (-1)^n a_n\) converges if: 1. \(\lim_{n\to\infty} a_n = 0\) 2. The sequence \(a_n\) is decreasing Now, let's check whether our given series satisfies these conditions.

Step 3: Find the limit of \(a_n\)

Let's find the limit of the sequence \(a_n = \frac{1}{\ln n}\) as \(n\) approaches infinity. \[ \lim_{n\to\infty} \frac{1}{\ln n} \] Since the natural logarithm function \(\ln n\) increases without bound as \(n \to \infty\), we know that its reciprocal \(\frac{1}{\ln n}\) will approach 0. Therefore: \[ \lim_{n\to\infty} \frac{1}{\ln n} = 0 \] The first condition of the Alternating Series Test is satisfied.

Step 4: Check if the sequence is decreasing

Now, we need to check if the sequence \(a_n = \frac{1}{\ln n}\) is decreasing. We'll do this by examining the derivative of \(a_n\). We have: \[ a_n = \frac{1}{\ln n} \] We can rewrite this as: \[ a_n = (\ln n)^{-1} \] Now, let's find the derivative of \(a_n\): \[ a_n' = -(\ln n)^{-2} \cdot \frac{1}{n} \] Since both terms in the product are positive for \(n\geq 2\), their product will be negative. Thus, the derivative is negative for \(n\geq 2\), which means that the sequence is decreasing. Therefore, the second condition of the Alternating Series Test is also satisfied.

Step 5: Conclusion

Since both conditions of the Alternating Series Test are satisfied, we can conclude that the alternating series \(\sum_{n=2}^{\infty} \frac{(-1)^{n-1}}{\ln n}\) converges.

Key Concepts

Convergence of Series

Understanding the convergence of series is essential in mathematics, especially when dealing with infinite sums. A series is said to converge if the sum of its infinite terms approaches a specific, finite value. This is a crucial concept because it helps to determine whether an infinite sum has a limiting value or simply grows without bounds.

When we look at the convergence of a series, we typically use specific tests based on the nature of the series. Some of these tests include the Alternating Series Test, the Ratio Test, the Root Test, and others. For instance, in the case of the given exercise, the Alternating Series Test is used to determine convergence. To pass this test, a series must have terms that tend to zero as well as decreasing absolute values of its successive terms.

It's essential to understand not just whether a series converges, but also how quickly it reaches its limit. This can help in approximating the sum of the series or understanding its behavior within certain mathematical models.

Alternating Series

An alternating series is a special type of series where its terms alternate in sign. This means that consecutive terms in the series will switch between positive and negative. For instance, a series like \(-1, 1, -1, 1, \textellipsis\) exhibits this alternating pattern.

An example of an alternating series in mathematical form would be \(\textstyle\frac{1}{\textstyle\frac{1}{\textln n}}\) where the minus sign is raised to an expression involving \(n\). Such series often converge due to the fact that the signed terms start to cancel each other out as the series progresses, which can lead to a stable sum despite the infinite number of terms.

It's worth noting that just because a series is alternating, it does not automatically mean it converges. The Alternating Series Test, as demonstrated in the solution steps, helps to confirm or deny convergence by checking specific conditions.

Series and Sequences

Series and sequences are fundamental concepts in calculus and mathematical analysis. A sequence is an ordered list of numbers, whereas a series is the sum of the terms of a sequence. It can be finite or infinite, and understanding the behavior of sequences is key to analyzing series.

In the context of the exercise shown, the sequence \(\textstyle\frac{1}{\textln n}\) forms the basis of the series. Studying sequences involves looking at properties such as convergence, bounds, and monotonicity. A sequence converging to a limit means that its terms can get as close as we desire to a specific value as the sequence progresses.

Furthermore, if a sequence is monotonic, meaning consistently increasing or decreasing, and is bounded, meaning its terms stay within a specific range, this can often lead to convergence. These concepts help in analyzing series and determining their long-term behavior, like whether or not their sums converge.

Natural Logarithm

The natural logarithm, denoted as \(\ln\), is a mathematical function that's fundamental in calculus. It's the inverse operation of exponentiation with the number \(e\) as the base, where \(e\) is an irrational and transcendental number approximately equal to 2.71828.

The natural logarithm has the property of transforming the multiplicative processes into additive ones, which is extremely useful in solving complex mathematical problems, especially those involving growth and decay processes. In the example provided, we encounter the natural logarithm in the denominator of the sequence, \(\textstyle\frac{1}{\textln n}\).

As \(n\) becomes large, the value of \(\ln n\) also grows without bounds. However, its growth is relatively slow compared to exponential growth, which is characterized by the function \(e^n\). This slow growth rate is what ultimately leads to the reciprocal \(\textstyle\frac{1}{\textln n}\) tending to zero as \(n\) approaches infinity, a critical point in proving the convergence of the series using the Alternating Series Test.