Question

Let \(S_{n}\) be the \(n^{\text {th }}\) partial sum of a series. In Exercises \(21-24,\) a convergent alternating series is given and a value of \(n .\) Compute \(S_{n}\) and \(S_{n+1}\) and use these values to find bounds on the sum of the series. $$\sum_{n=0}^{\infty}\left(-\frac{1}{2}\right)^{n}, \quad n=9$$

Short answer

Calculate partial sums: \(S_9 = \frac{683}{1024}\) and \(S_{10} = \frac{683}{1024}\); sum bounds: \( \frac{341}{512} < S < \frac{171}{256} \).

Step-by-step solution

Understanding the Series

The given series is an infinite geometric series with terms of the form \( a_n = \left(-\frac{1}{2}\right)^n \). This series is alternating because each term switches sign based on \((-1)^n\). The formula for the sum of an infinite geometric series \( \sum a r^n \) is \( \frac{a}{1-r} \) when \(|r| < 1\), where \(a\) is the first term \( (a_0=1) \) and \(r\) is the common ratio \(-\frac{1}{2}\).

Calculate S9 (Partial Sum to n = 9)

The \(n^{th}\) partial sum for a geometric series \( S_n \) can be calculated using \( S_n = \frac{a(1-r^{n+1})}{1-r} \). For our series, \(a = 1\) and \(r = -\frac{1}{2}\), so:\[ S_9 = \frac{1(1-(-\frac{1}{2})^{10})}{1 - (-\frac{1}{2})} = \frac{1 - \frac{1}{1024}}{\frac{3}{2}} = \frac{1023}{1024} \cdot \frac{2}{3} = \frac{2046}{3072} = \frac{683}{1024} \]

Calculate S10 (Partial Sum to n = 10)

Similarly, calculate \( S_{10} \) using the formula:\[ S_{10} = \frac{1(1-(-\frac{1}{2})^{11})}{1 - (-\frac{1}{2})} = \frac{1+\frac{1}{2048}}{\frac{3}{2}} = \frac{2049}{2048} \cdot \frac{2}{3} = \frac{4098}{6144} = \frac{1366}{2048} = \frac{683}{1024} \]

Determine Bounds on the Sum of the Series

In an alternating series, the error when using the \(n^{th}\) partial sum to approximate the total sum is less than the first omitted term. After \(n = 9\), the first omitted term is \(\left|-\frac{1}{2}\right|^{10} = \frac{1}{1024}\). This means the true sum of the series, \(S\), is bounded by:\[ S_9 - \left| (-\frac{1}{2})^{10} \right| < S < S_9 + \left| (-\frac{1}{2})^{10} \right| \] \[ \frac{683}{1024} - \frac{1}{1024} < S < \frac{683}{1024} + \frac{1}{1024} \] \[ \frac{682}{1024} < S < \frac{684}{1024} \] Which simplifies to:\[ \frac{341}{512} < S < \frac{171}{256} \]

Key concepts

Geometric Series

A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. In the context of the provided problem, the series seems to take the form of \( \sum \left(-\frac{1}{2}\right)^{n} \), where each term is obtained by multiplying the preceding term by \(-\frac{1}{2}\).
This makes it not just a geometric series, but an **alternating** geometric series because the sign of each term flips (+, -, +, -) as a result of the negative ratio. The common ratio \( r \) here is \(-\frac{1}{2}\). - Understanding this characteristic makes it easier to analyze and predict the behavior of the series in terms of its eventual convergence or divergence.
The infinite geometric series can be summed if the absolute value of the common ratio \(|r|\) is less than 1. The formula to find this sum is \( \frac{a}{1-r} \), where \( a \) is the first term of the series. This helps calculate the total sum if the series is infinitely long.

Partial Sum

The concept of partial sums is critical for evaluating series since it gives a way to approximate the total sum of an infinite series using just a few terms. In the exercise, when calculating the 9th partial sum \( S_9 \), we've summed the first nine terms of the series.
The formula for finding the partial sum, \( S_n \), of a geometric series is \(S_n = \frac{a(1 - r^{n+1})}{1 - r} \).- This formula ensures you are not just analyzing each term individually but rather how they accumulate up to a certain point. By computing \( S_9 \) and \( S_{10} \), you can see how close these sums get to the actual sum of the infinite series, highlighting the series' convergence rate. It's a manageable way to understand infinite sums by breaking them down into finite pieces.

Series Convergence

Convergence in series determines whether the sum of infinite terms approaches a finite number, as is the case with the alternating series here. - For geometric series with a common ratio's absolute value less than 1, such as our series with \( r = -\frac{1}{2} \), convergence is guaranteed. The alternating nature means each term in the series reduces in size, rendering progressively smaller contributions to the total sum. The convergence happens due to the balancing act of adding positive and negative terms which decreases the overall sum fluctuations.For the given series, the convergence was //explored// by looking at the value of \( n \) for computing \( S_9 \) and \( S_{10} \) to represent the behavior closely approaching a sum. This inherent feature is foundational for understanding infinite series as many similarly structured series share this property.

Error Bound

In evaluating infinite series, particularly alternating ones, the error bound gives insight into how accurate your approximation is. The error bound refers to the difference between the actual sum of the series and the partial sum \( S_n \).For alternating series, the error after \( n \) terms can be assessed by the first omitted term, making it an easily manageable task. - This means that for the given series, the omitted term after the 9th partial sum is \( \left|\left(-\frac{1}{2}\right)^{10}\right| = \frac{1}{1024} \).By employing this, you get a numeric range \( S_9 - \epsilon < S < S_9 + \epsilon \) (where \( \epsilon \) is the omitted term) that we can use to denote how close our partial sum is to the actual sum. This use of error bounds demonstrates a practical technique for estimating the limits of infinite series when making direct calculation isn't feasible.