91Ó°ÊÓ

In mathematics, a series is a sum of sequences of terms. When we discuss the convergence of a series, we are interested in whether the sum of these infinitely many terms approaches a finite number. With geometric series, the focus is often on finding conditions where this sum stabilizes rather than blowing up to infinity. For a geometric series to converge, the absolute value of the common ratio, denoted by \( |r| \), must be less than 1. This specific condition ensures that the terms become smaller and smaller, allowing their sum to approach a fixed value.
This is crucial because if \( |r| \geq 1 \), the terms do not shrink and the series may diverge, meaning it doesn't sum to a finite limit.

The interval of convergence refers to the range of values for the variable in a series that allow the series to converge. For a geometric series given by \( \, \sum_{n=0}^{\infty} ar^n \), the interval is determined by solving the inequality \( |r| < 1 \).

In our exercise, the common ratio \( r \) is \((-4)(x-5)\). We set up the inequality
\[4|x-5| < 1\]
which simplifies to \(|x-5| < \frac{1}{4}\). This inequality tells us that for the series to converge, \( x \) must lie within a specific range.

• First, rewrite the inequality as \(-\frac{1}{4} < x-5 < \frac{1}{4}\).
• Solving this gives \( \frac{19}{4} < x < \frac{21}{4} \).

This means the interval of convergence for \( x \) is between \( \frac{19}{4} \) and \( \frac{21}{4} \).

Once a geometric series is determined to be convergent, we can find its sum. The formula for the sum of a convergent geometric series is given by \( S = \frac{a}{1-r} \), where \( a \) is the first term of the series, and \( r \) is the common ratio.

In the exercise, the series is expressed as \( \, \sum_{n=0}^{\infty} (-4)^{n}(x-5)^{n} \). The first term \( a \) is \( 1 \) when \( n = 0 \), and the common ratio \( r \) is \((-4)(x-5)\).
Therefore, the sum \( S \) is:
\[S = \frac{1}{1-(-4)(x-5)} = \frac{1}{1 + 4(x-5)}\]
Simplifying the expression leads to:
\[S = \frac{1}{4x - 19}\]
This sum is valid for the values of \( x \) defined in the interval of convergence \( \left( \frac{19}{4}, \frac{21}{4} \right) \).

Solving absolute value inequalities is a key part of finding the interval of convergence. It entails determining where a given expression involving an absolute value is either less than, greater than, or equal to a particular number.

For the exercise, the inequality \(|-4(x-5)| < 1\) needed to be solved. Here's how:

• First, simplify by removing the factor of \(-4\):\(|x-5| < \frac{1}{4}\).
• This means the expression inside the absolute value, \(x-5\), is restricted to lie between \(-\frac{1}{4}\) and \(\frac{1}{4}\). Hence, \(-\frac{1}{4} < x-5 < \frac{1}{4}\).

To eliminate the absolute value, split this inequality into two separate inequalities:
\[-\frac{1}{4} < x-5\] and \[ x-5 < \frac{1}{4} \].
Solve each to find the bounds for \( x \):
\(-\frac{1}{4} < x-5\) transforms to \(x > \frac{19}{4}\), while \(x-5 < \frac{1}{4}\) yields \(x < \frac{21}{4}\).
Thus, together they define the interval where the series converges: \( \frac{19}{4} < x < \frac{21}{4} \). This method creates the necessary range for \( x \) without altering the problem's context.