Problem 26 - Single Variable Calculus

Question

Find the values of \(p\) for which the series is convergent. $$ \sum_{n=2}^{\infty}(-1)^{n-1} \frac{(\ln n)^{p}}{n} $$

Step-by-Step Solution

Answer

The series \(\sum_{n=2}^{\infty}(-1)^{n-1} \frac{(\ln n)^{p}}{n}\) converges for \(p > 0\).

Step 1: Split the series into alternating part and positive term

First, we split the given series into the alternating part and the positive term, i.e., we have \((-1)^{n-1}\) as the alternating part and \(\frac{(\ln n)^{p}}{n}\) as the positive term \(a_n\): $$ \sum_{n=2}^{\infty}(-1)^{n-1} \frac{(\ln n)^{p}}{n} $$

Step 2: Check for descending sequence of positive terms

In order to apply the Alternating Series Test, we need to check if the positive term sequence is descending, i.e., if \(a_n \geq a_{n+1}\) for all \(n\). We calculate the derivative of the term \(\frac{(\ln n)^{p}}{n}\), to verify if it is decreasing: Let $$ g(n) = \frac{(\ln n)^{p}}{n} $$ We want to study when the derivative of \(g(n)\) is less than zero: $$ g'(n) = \frac{d}{dn}\left(\frac{(\ln n)^{p}}{n}\right) $$ Using the quotient rule of differentiation, we have: $$ g'(n) = \frac{n(\ln n)^{p-1} - (\ln n)^{p}}{n^2} $$ Now, we are interested in finding conditions for which \(g'(n) < 0\).

Step 3: Check the convergence condition

To find the convergent series, we need to know when \(\lim_{n\to\infty} a_n = 0\). In other words, we want to find for which values of \(p\), we have: $$ \lim_{n\to\infty} \frac{(\ln n)^{p}}{n} = 0 $$ Since the natural logarithm function increases slower than the linear function when \(n \to \infty\), the limit will be zero when \(p > 0\).

Step 4: Combine the results

Based on the analysis in the previous steps, the series converges when: 1. The positive terms are decreasing, i.e., \(g'(n) < 0\). 2. The limit of the positive terms goes to zero when \(n \to \infty\). From our findings, we can conclude that the series converges for \(p > 0\).

Key Concepts

Alternating Series Test

The Alternating Series Test is a method used to determine whether an infinite series with an alternating sign converges. When you encounter a series where the terms are switching between positive and negative, it's like a signal that this test might be relevant. For the test to be applicable, two conditions must be met:

• The absolute value of the terms must decrease monotonically, which means each term is less than or equal to the previous term in absolute value.
• The limit of the terms as they approach infinity must be zero.

If both conditions are fulfilled, we can definitely say that the series converges. This test is particularly handy because it doesn't just tell us if the sum is finite; it also reassures us that partial sums of the series will eventually settle down close to some fixed number as more and more terms are added.

Series Convergence

Understanding series convergence is essential for exploring infinite series, such as the one given in our exercise. A series is said to converge if the sum of its terms approaches a specific, finite value as the number of terms increases indefinitely. If there's no such limit, the series diverges. It's like filling up a glass of water: if you can actually fill it to the brim (finite value), the series converges; if the water keeps spilling out without ever filling up, that's divergence. To check for convergence, mathematicians have developed multiple tests, like the Alternating Series Test, and it all comes down to checking the behavior of the series' terms as they grow without bound. This concept is fundamental in calculus and analysis, since many problems involve summing an infinite number of terms—for example, in determining the value of certain functions or solving differential equations.

Natural Logarithm

The natural logarithm, denoted as \(\ln\), is a mathematical function that’s the inverse of the exponential function with base \(e\), where \(e\) is an irrational and transcendental constant approximately equal to 2.71828. In simpler terms, the natural logarithm of a number \(x\) answers the question: 'To what power must we raise \(e\) to get \(x\)?' Its involvement in calculus is widespread, from integration to solving differential equations, because it has a derivative that's particularly easy to work with, namely \(1/x\). In our exercise, we're interested in how this function behaves as it's raised to a power \(p\) and then divided by \(n\) as \(n\) becomes very large. One key property is that \(\ln n\) grows much slower than \(n\), which is core to understanding why the provided series will converge for certain values of \(p\).

Quotient Rule of Differentiation

When dealing with the differentiation of a function that is the quotient of two other functions, the Quotient Rule comes into play. The rule states that to differentiate \(\frac{f(x)}{g(x)}\), you subtract the product of the derivative of the numerator \(f\) and the denominator \(g\) from the product of the numerator and the derivative of the denominator, all divided by the square of the denominator. It looks like this:
\[\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}\]This rule is a critical tool in calculus since it allows the differentiation of complex expressions where division is involved. In our problem, we used the Quotient Rule to help us find when the derivative of the term \(\frac{(\ln n)^p}{n}\) is negative, which in turn informs us about the decreasing nature of the series' terms—a vital step in deciding whether our series is convergent.