91Ó°ÊÓ

A geometric series is a series of the form \( a + ar + ar^2 + ar^3 + \ldots \) where \( a \) is the first term and \( r \) is the common ratio between the terms. A geometric series converges if the absolute value of the common ratio \( r \) is less than 1, i.e., \( |r| < 1 \). When it converges, the sum of the geometric series can be calculated using the formula:

• \( S = \frac{a}{1-r} \)

Let's apply this to a part of the original problem: the series \( \sum_{n=0}^{\infty}5^{n}(x-3)^{n} = \sum_{n=0}^{\infty}\left(5(x-3)\right)^n \). Here, the common ratio \( r \) becomes \( 5(x-3) \), and for convergence, we require \( |5(x-3)| < 1 \). Solving this inequality leads to the interval \( |x-3| < \frac{1}{5} \). Hence, this series converges absolutely when \( x \) is in the interval \( (\frac{14}{5}, \frac{16}{5}) \).
This principle of convergence of geometric series is crucial for understanding power series behaviors.

The exponential series is a special type of series that expands the exponential function \( e^x \). This series takes the form:

• \( \sum_{n=0}^{\infty}\frac{x^n}{n!} \)

The beautiful property of this series is that it converges for all values of \( x \), making it universally applicable. In our exercise, the series \( \sum_{n=0}^{\infty}\frac{4^{n}}{n!}(x-3)^{n} \) resembles the exponential series \( e^{4(x-3)} \). Here, since it keeps the same structure as the general exponential series, it will always converge, regardless of the value of \( x \). This universal convergence is because each term \( \frac{x^n}{n!} \) gets proportionally smaller as \( n \) increases, leading to a rapidly converging sum.

The interval of convergence is a range of \( x \) values where a power series converges absolutely. For most power series, determining this interval is crucial as it defines the values for which the series converges to a finite sum. To find the interval, we often analyze components of the power series separately. In our problem, the series was split into a geometric and an exponential component:

• The geometric component \( \sum_{n=0}^{\infty}5^{n}(x-3)^{n} \) converges if \( |x-3| < \frac{1}{5} \).
• The exponential component \( \sum_{n=0}^{\infty}\frac{4^{n}}{n!}(x-3)^{n} \) converges for all \( x \).

Given these, the interval of convergence for the entire series is dictated by the geometric component. This means the series converges absolutely within \( x \) values from \( (\frac{14}{5}, \frac{16}{5}) \). The interval of convergence provides insight into where the power series can sensibly be used to approximate functions. Understanding this helps in practical applications and deeper mathematical concepts.