91Ó°ÊÓ

An infinite series is the sum of an infinite sequence of terms. In this problem, the series is shown as a sum of several terms like \( 1 - \frac{1}{2}(x-3) + \frac{1}{4}(x-3)^{2} + \cdots \). Each term is a product of a constant factor and a power of \((x - 3)\). In simple terms, the infinite series continues indefinitely. It doesn't stop after a few terms, which is why understanding its behavior over an infinite number of terms becomes essential. Infinite series are crucial in calculus because they help in representing complex mathematical functions in a manageable way.
Some key points about infinite series include:

• They involve the addition of infinitely many terms.
• The terms can be derived using a specific formula.
• They may converge to a certain value, or diverge without reaching any limit.

Understanding how these series behave as the number of terms goes to infinity helps in analyzing their convergence or divergence.

Convergence of an infinite series refers to whether the sum of its terms approaches a finite limit as more and more terms are added. For the series given in the exercise, it converges under certain conditions. Specifically, it's a geometric series described by a common ratio. A geometric series will converge if the absolute value of the common ratio \(|r|\) is less than 1.
In our problem, the common ratio is \(-\frac{1}{2}(x-3)\), and for convergence, we ensure \(\left| -\frac{1}{2}(x-3) \right| < 1\). Solving this inequality, we find the series converges when \(1 < x < 5\).
Here's a quick checklist for determining convergence of a geometric series:

• Identify the first term and common ratio.
• Check if \(|r| < 1\). If yes, the series converges.
• The convergence of other series types might require different techniques.

So, the series converges for values of \(x\) in the interval \((1, 5)\).

Differentiation involves finding the derivative of a function, which measures how the function changes as its input changes. In the context of series, differentiating term by term is a useful technique, especially for power series, like in this exercise.
Differentiating each term \(a_n = \left(-\frac{1}{2}\right)^n (x-3)^n\) gives us the derivative:\( \frac{d}{dx}(a_n) = n \left(-\frac{1}{2}\right)^n (x-3)^{n-1} \). This generates a new series starting from \(n = 1\) instead of \(n = 0\) since the first constant term's derivative is zero. This operation is possible because the original series can be seen as an infinite polynomial summed across terms.
Here are some steps for differentiating series term by term:

• Differentiation should respect the power of \(x\) terms.
• Start differentiation for terms where \(n \geq 1\).
• Check for any changes in convergence behavior post-differentiation.

After differentiation, it is important to re-evaluate the interval of convergence as it might remain the same or change.

An interval of convergence is the range of values for which an infinite series converges. For both the original and differentiated series, the interval of convergence is the same, \((1, 5)\).
Determining the interval of convergence is crucial for working with series, as it is only within this interval that the series makes sense and converges to a specific value.
Key considerations for finding the interval of convergence include:

• Solve inequalities for the absolute value of the common ratio.
• Identify the endpoints of the interval by checking boundaries.
• Use these to define whether it converges to infinity or a real number.

The interval of convergence tells when the terms add up to form a coherent limit and is vital for ensuring the series’ utility in approximations and calculations.