Question

Find the function represented by the following series and find the interval of convergence of the series. $$\sum_{k=0}^{\infty}(\sqrt{x}-2)^{k}$$

Short answer

Answer: The function represented by the given series is $f(x) = \frac{1}{3-\sqrt{x}}$. The interval of convergence of the series is $1 < x < 9$.

Step-by-step solution

Identify the geometric series and its parameters

The given series is: $$\sum_{k=0}^{\infty}(\sqrt{x}-2)^{k}$$ This is a geometric series with the first term, \(a = 1\), and the common ratio, \(r = (\sqrt{x}-2)\).

Find the sum of the geometric series

To find the sum of the geometric series, use the formula for the sum of an infinite geometric series: $$S = \frac{a}{1-r}$$ Plugging in the first term, \(a\), and the common ratio, \(r\), we have: $$S = \frac{1}{1-(\sqrt{x}-2)}$$

Simplify the sum

Now, simplify the sum: $$S = \frac{1}{1-\sqrt{x}+2}$$ $$S = \frac{1}{3-\sqrt{x}}$$ So the function represented by the given series is: $$f(x) = \frac{1}{3-\sqrt{x}}$$

Find the radius and interval of convergence

For a geometric series to converge, the absolute value of the common ratio must be less than 1, i.e., \(|r| < 1\). In this case, our common ratio is \((\sqrt{x}-2)\). Apply the convergence condition: $$|(\sqrt{x}-2)| < 1$$ Now, solve the inequality for \(x\): $$-1 < (\sqrt{x}-2) < 1$$ $$1 < \sqrt{x} < 3$$ Square all the parts of the inequality: $$1^2 < (\sqrt{x})^2 < 3^2$$ $$1 < x < 9$$

Write the final answer

The function represented by the given series is: $$f(x) = \frac{1}{3-\sqrt{x}}$$ The interval of convergence of the series is: $$1 < x < 9$$

Key concepts

Series Convergence

In mathematics, a series is said to converge when its terms approach a specific value as more and more terms are added. For a geometric series to converge, a crucial requirement is that the absolute value of its common ratio should be less than 1. This condition ensures that the infinite sum will approach a finite limit.

A geometric series can be expressed as:

• The first term, usually denoted as \(a\).
• A common ratio, \(r\), multiplied to the succeeding terms.

The series becomes: \(a, ar, ar^2, ar^3, \ldots\). As these terms are summed indefinitely, convergence depends wholly on the common ratio \(r\). For our example given by the exercise, the series \( \sum_{k=0}^{\infty} (\sqrt{x}-2)^{k} \) is a convergence checking playground.

Since convergence requires \(|r| < 1\), here \(r = (\sqrt{x} - 2)\). By solving this inequality, you determine whether or not the series sums to a finite number.

Radius of Convergence

The radius of convergence is a fundamental concept when working with power series. It marks the distance from the center of the series to the points at which the series stops converging. In simpler terms, this radius helps identify how far away from a chosen center we can go before the series becomes divergent.

The radius of convergence can be found using the formula:

• \( |r| = \text{smallest positive solution of the inequality } |r| < 1 \)

In a geometric series such as \( \sum_{k=0}^{\infty} (\sqrt{x} - 2)^k \), the radius equates to determining when \( |\sqrt{x} - 2| < 1 \). From the solution process:

• This simplifies to the condition \( 1 < x < 9 \)

The radius of convergence then corresponds to half of the range of \(x\), or \(4\), effectively capturing how wide the interval is when centered around \((\sqrt{1} + \sqrt{9}) / 2 = 2\). This tells us that within this radial range, the series maintains its ground firmly on convergence.

Interval of Convergence

An interval of convergence refers to the range or span of variable values for which a series converges. Oftentimes, particularly in geometric or power series, finding this interval is critical to understanding where the function behaves properly.

In the given exercise, we derived the interval for \(x\) by solving \(| \sqrt{x} - 2 | < 1\). This resulted in:

• The inequality \( 1 < \sqrt{x} < 3 \).

Upon squaring each part of the inequality, we determined

• The interval \( 1 < x < 9 \).

This means that only within this interval will the series converge to a finite sum. If you choose an \(x\) value outside of this range, the series would diverge, meaning it won’t sum to a meaningful, finite number. Exploring the interval shows how sensitive functions represented by infinite series, like our function \(f(x) = \frac{1}{3 - \sqrt{x}}\), are to the specific values they are assigned.