Problem 39 - Calculus

Question

By recognizing each series as a Taylor series evaluated at a particular value of \(x,\) find the sum of each of the following convergent series. $$\frac{1}{2}-\frac{\left(\frac{1}{2}\right)^{2}}{2}+\frac{\left(\frac{1}{2}\right)^{3}}{3}-\frac{\left(\frac{1}{2}\right)^{4}}{4}+\cdots+\frac{(-1)^{n} \cdot\left(\frac{1}{2}\right)^{n+1}}{(n+1)}+\cdots$$

Step-by-Step Solution

Answer

The sum is \( \ln\left(\frac{3}{2}\right) \).

Step 1: Identify the Series

The given series is \( \frac{1}{2} - \frac{\left(\frac{1}{2}\right)^{2}}{2} + \frac{\left(\frac{1}{2}\right)^{3}}{3} - \frac{\left(\frac{1}{2}\right)^{4}}{4} + \cdots \). This series closely resembles the Taylor series expansion for \( \ln(1 - x) \), which is \( -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \cdots \), evaluated at a particular value of \( x \).

Step 2: Determine the Function

Notice the alternating nature (-1) of our series and compare it to the known Taylor series: \( \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots \). Our series matches with \( \ln(1+x) \), setting \( x = \frac{1}{2} \).

Step 3: Evaluate the Function at Specific x Value

Now that we identify the series as the expansion of \( \ln(1+x) \) where \( x = \frac{1}{2} \), we can evaluate: \( \ln(1+\frac{1}{2}) = \ln\left(\frac{3}{2}\right) \).

Step 4: Calculate the Sum

The sum of our series is given by the evaluation of the function. Hence, the series converges to \( \ln\left(\frac{3}{2}\right) \).

Key Concepts

Convergent Series

A convergent series is one where the sum of its terms gets closer and closer to a specific number as more terms are added. Imagine continuing a sum without it heading off to infinity or never settling down to a particular value. That's convergence!

Let's take the series given in the exercise as an example, which is:

• It starts with fractions: \( \frac{1}{2} - \frac{\left(\frac{1}{2}\right)^{2}}{2} + \frac{\left(\frac{1}{2}\right)^{3}}{3} - \cdots \)

Here, the positive and negative terms hint towards eventual convergence. The alternating nature, i.e., the changing of positive and negative signs, often helps a series in converging by balancing the evens and odds.

• In this series, not only do signs alternate, but even the magnitude of each term decreases steadily.
• This is due to the powers of \( \frac{1}{2} \) getting smaller as the series progresses.

Each part's successive smaller size and alternating nature help our series converge, finally allowing us to find a particular value at which it levels out.

Logarithmic Functions

Logarithmic functions are the opposite of exponential functions. The logarithm measures the time it takes for a certain exponential growth to reach a particular level. When you think about logarithms, think about a tree developing at a steady pace, but your growth checks only show at certain intervals instead of continuous updates.

The focus in this context is the natural logarithm, denoted by \( \ln(x) \). This is the logarithm to the base \( e \), where \( e \) (approximately 2.718) is a special constant. It's commonly used in calculus due to its natural appearance in various growth processes, like populations in nature or interest calculations.

In the original exercise, the series reflects the Taylor series expansion for the natural logarithm, specifically \( \ln(1+x) \). Taylor series helps us express complex functions, like \( \ln \), into simpler power series, making calculations more straightforward.

• For a small \( x \), \( \ln(1+x) \) can be approximated as: \[ x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots \]

Essentially, by using this expansion, the original problem turned into evaluating \( \ln(1+x) \) at \( x=\frac{1}{2} \), showing how logarithmic functions help us solve the problem.

Series Evaluation

Series evaluation is about finding the total or final value of a series after identifying its pattern or rule. It's like putting a puzzle together piece by piece until you see the whole picture. In our exercise, this evaluation happens through recognizing patterns in the series.

Initially, we identified the significance of each term aligning with a known Taylor series, allowing us to improve evaluation methods, rather than adding each term laboriously:

• The series given: \( \frac{1}{2} - \frac{\left(\frac{1}{2}\right)^{2}}{2} + \frac{\left(\frac{1}{2}\right)^{3}}{3} - \cdots \)
• This forms a known expansion derivative of the Taylor series for \( \ln(1+x) \), evaluated at \( x=\frac{1}{2} \).
• Upon recognition, we determined that the entire series evaluates to \( \ln\left(\frac{3}{2}\right) \).

Evaluating such a series involves understanding patterns and properties of the series so that we can sum up infinitely many numbers easily. This requires both intuition and structured mathematical reasoning to solve and verify correctly.