Problem 111 - College Algebra with Modeling and Visualization

Question

Use \(a_{k}\) and \(n\) to find \(S_{n}=\Sigma_{k-1}^{n} a_{k}\) (Refer to Example \(6 .\) ) Then evaluate the infinite geometric series \(S=\sum_{k=1}^{\infty} a_{k},\) Compare \(S\) to the values for \(S_{n^{}}.\) $$a_{k}=4\left(-\frac{1}{10}\right)^{k-1} ; n=1,2,3,4,5,6$$

Step-by-Step Solution

Answer

The infinite sum is \(\frac{40}{11}\), and \(S_n\) values approach it as \(n\) increases.

Step 1: Understand the Problem

We need to sum the series given by \(a_k = 4\left(-\frac{1}{10}\right)^{k-1}\) from \(k=1\) to \(n\) for different values of \(n\) and also evaluate the infinite series.

Step 2: Apply Formula for Finite Geometric Series

The sum of the first \(n\) terms of a geometric series is given by \(S_n = \frac{a_1(1-r^n)}{1-r}\), where \(a_1 = 4\) and \(r = -\frac{1}{10}\).

Step 3: Calculate Each Term for Given Values of n

Substitute \(a_1 = 4\) and \(r = -\frac{1}{10}\) into the formula. Compute \(S_n\): - For \(n = 1\), \(S_1 = \frac{4(1-(-\frac{1}{10}))}{1-(-\frac{1}{10})} = 4\) - For \(n = 2\), \(S_2 = \frac{4(1-(-\frac{1}{10})^2)}{1-(-\frac{1}{10})}\) - Continue with \(n = 3, 4, 5, 6\).

Step 4: Evaluate the Infinite Series

Use the infinite geometric series formula \(S = \frac{a_1}{1-r}\). Since \(r = -\frac{1}{10}\), \(|r| < 1\), so the series converges. Calculate \(S = \frac{4}{1 - (-\frac{1}{10})} = \frac{4}{1 + \frac{1}{10}} = \frac{4}{1.1} = \frac{40}{11}\).

Step 5: Compare Finite and Infinite Series Sums

The sums \(S_n\) are calculated for each \(n\) and should approach \(S\) as \(n\) increases. By comparing the calculated \(S_n\) with \(\frac{40}{11}\), observe the convergence.

Key Concepts

Finite Geometric Series

A finite geometric series is a sum of terms in a geometric sequence that has a limited number of terms. Imagine you are adding a certain number of terms together, starting with a first term and multiplying by a constant factor known as the common ratio. For a series like this, the sum of the first \(n\) terms can be calculated by using the formula: \[ S_n = \frac{a_1(1-r^n)}{1-r} \] where:

• \(a_1\) is the first term of the series.
• \(r\) is the common ratio (the fixed number each term is multiplied by to get the next term).
• \(n\) is the number of terms you are summing up.

Using this formula, each term can be computed by substituting the appropriate values. For the series given in the exercise, with \(a_k = 4\left(-\frac{1}{10}\right)^{k-1}\),\(a_1 = 4\), and \(r = -\frac{1}{10}\), we can calculate \(S_n\) for different values of \(n\) like 1, 2, 3, etc. Each computed sum takes into account the decreasing magnitude of each subsequent term, as they are affected by the small ratio \(-\frac{1}{10}\). As \(n\) increases, we'll notice the sum of these terms converges towards a specific value, as we explore next!

Infinite Series

An infinite geometric series extends indefinitely, adding an unlimited number of terms. This type of series only converges (approaches a specific value) when the absolute value of the common ratio \(|r|\) is less than 1. For our series, the common ratio \(-\frac{1}{10}\) meets this criterion, meaning the series converges to a finite sum even though there are infinitely many terms. The sum of an infinite geometric series is determined using the formula: \[ S = \frac{a_1}{1-r} \] For our given series: - \(a_1 = 4\)- \(r = -\frac{1}{10}\) Using the formula, we calculate: \[ S = \frac{4}{1 - (-\frac{1}{10})} = \frac{4}{1 + \frac{1}{10}} = \frac{4}{1.1} = \frac{40}{11} \] This result represents the ultimate sum of the series as the number of terms approaches infinity.

Series Convergence

Series convergence describes the behavior of a series as more and more terms are added. Essentially, it's about understanding whether a series will settle towards a particular value or diverge into infinity. In our example, the series converges since the common ratio has an absolute value less than one. Convergence is important because it allows us to find an actual sum for a series that theoretically never ends.

• If \(|r|<1\), like in our exercise, the series will converge to \(\frac{a_1}{1-r}\).
• If \(|r| \geq 1\), the series diverges, meaning it does not approach any finite value.

Comparing finite sums \(S_n\) with the infinite sum \(\frac{40}{11}\) shows how close each finite sum gets to this ultimate total as \(n\) increases. This illustrates convergence in action, with each finite sum bringing us closer to the infinite series' resolved value. The beauty of convergence lies in the stability and certainty it provides to our understanding of infinite processes.