Problem 87 - A Graphical Approach to Precalculus with Limits

Question

Determine the positive integer values of \(n\), where the sequence \(a_{n}\) satisfies the inequality. $$a_{n} \geq \frac{1}{5}, \text { where } a_{n}=\frac{n}{1+n^{2}}$$

Step-by-Step Solution

Answer

The positive integer values are \(n = 1, 2, 3, 4\).

Step 1: Setup the Inequality

The problem requires finding the integer values of \(n\) for which \(a_n \geq \frac{1}{5}\). Start by substituting \(a_n = \frac{n}{1+n^2}\) into this inequality: \[ \frac{n}{1+n^2} \geq \frac{1}{5}. \]

Step 2: Clear the Fractions

To eliminate the fractions, multiply both sides of the inequality by \(5(1+n^2)\) to avoid any division by zero:\[ 5n \geq 1 + n^2. \]

Step 3: Rearrange Terms

Bring all terms to one side of the inequality to form a standard quadratic inequality:\[ n^2 - 5n + 1 \leq 0. \]

Step 4: Solve the Quadratic Equation

Find the roots of the quadratic equation \(n^2 - 5n + 1 = 0\) using the quadratic formula:\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \]where \(a = 1\), \(b = -5\), and \(c = 1\):\[ n = \frac{5 \pm \sqrt{25 - 4}}{2} = \frac{5 \pm \sqrt{21}}{2}. \]

Step 5: Approximate the Roots

Calculate the approximate values of the roots:\[ n = \frac{5 + \sqrt{21}}{2} \approx 4.79, \quad n = \frac{5 - \sqrt{21}}{2} \approx 0.21. \]

Step 6: Determine Integer Solutions

The quadratic inequality \(n^2 - 5n + 1 \leq 0\) has solutions for \(n\) between the roots \(n \approx 0.21\) and \(n \approx 4.79\). The integer solutions are \(n = 1, 2, 3, 4\), as these are the positive integers within this range.

Key Concepts

Quadratic Inequalities

Quadratic inequalities involve expressions where the unknown variable is squared. These inequalities are essential in problems where one needs to determine a range of values that meet certain conditions. In this context, we are dealing with an inequality arising from a sequence expression, which we need to solve for integer values of the variable \(n\).In the given problem, we begin with the inequality \(\frac{n}{1+n^2} \geq \frac{1}{5}\). This is a fractional expression, but by carefully manipulating it, we can transform it into a standard quadratic form. The inequality is simplified by multiplying through by \(5(1+n^2)\) to eliminate the fraction, leading to the quadratic inequality \(n^2 - 5n + 1 \leq 0\). This step is crucial, as it allows us to focus solely on the quadratic expression, paving the path for further solution steps. Understanding how to manage the transition from a fractional inequality to a standard quadratic one is key when handling such problems. It ensures that we systematically approach the solution by first clearing complex fractions, which often intimidate students.

Quadratic Formula

The quadratic formula is a reliable method for solving quadratic equations. It works universally for any quadratic equation of the form \(ax^2 + bx + c = 0\). In our exercise, the quadratic inequality \(n^2 - 5n + 1 \leq 0\) becomes an equation \(n^2 - 5n + 1 = 0\) for which we need the roots.For any quadratic equation \(ax^2 + bx + c = 0\), the quadratic formula is:

• \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

In this exercise, \(a = 1\), \(b = -5\), and \(c = 1\). Plugging these values into the quadratic formula gives the roots:

• \(n = \frac{5 \pm \sqrt{25 - 4}}{2}\)
• \(n = \frac{5 \pm \sqrt{21}}{2}\)

The roots are approximately \(0.21\) and \(4.79\). Finding these roots is essential, as they define the boundaries within which the quadratic inequality holds true. The quadratic formula saves time and simplifies complex calculations that might otherwise rely on trial and error.

Inequality Solutions

Solving inequalities, particularly quadratic ones, involves determining the range within which the inequality holds. From the quadratic inequality \(n^2 - 5n + 1 \leq 0\), we use the roots derived from the quadratic formula to explore possible solutions for \(n\).The critical step is recognizing that for \(n^2 - 5n + 1 \leq 0\), integer solutions must lie between the approximate roots \(0.21\) and \(4.79\). Thus, the integer values satisfying this inequality are \(n = 1, 2, 3, 4\).Why only these integers?

• Quadratic inequalities
  • tend to switch signs at their roots, indicating where the inequality will hold or not
• By identifying which sections of the number line satisfy the inequality, we select the relevant integers.

Understanding the behavior of quadratic inequalities through their roots and testing intervals is critical for success in solving these problems, as it directly informs us which integer values satisfy the given condition.