Question

A convergent alternating series is given along with its sum and a value of \(\varepsilon\). Use Theorem 8.5 .2 to find \(n\) such that the \(n^{\text {th }}\) partial sum of the series is within \(\varepsilon\) of the sum of the series. $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{4}}=\frac{7 \pi^{4}}{720}, \quad \varepsilon=0.001$$

Short answer

The smallest number of terms needed is \( n = 9 \).

Step-by-step solution

Understand the Alternating Series Remainder Theorem

Theorem 8.5.2 refers to the Alternating Series Remainder Theorem, which states that for a convergent alternating series \( \sum_{n=1}^{\infty} (-1)^{n+1} a_n \), the remainder \( R_n \) after \( n \) terms satisfies \( |R_n| = |S - S_n| \leq a_{n+1} \), where \( S \) is the total sum of the series and \( S_n \) is the sum of the first \( n \) terms.

Set Up Inequality for Remainder

To find \( n \) such that the \( n^{\text{th}} \) partial sum is within \( \varepsilon = 0.001 \) of \( S \), we use the inequality \( a_{n+1} \leq \varepsilon \). In our series, \( a_n = \frac{1}{n^4} \), thus \( a_{n+1} = \frac{1}{(n+1)^4} \leq 0.001 \).

Solve the Inequality

Now, solve \( \frac{1}{(n+1)^4} \leq 0.001 \). By taking reciprocals, we get \( (n+1)^4 \geq 1000 \). Taking the fourth root of both sides, \( n+1 \geq 10 \). Therefore, \( n \geq 9 \).

Determine the Smallest n

From \( n \geq 9 \), the smallest integer \( n \) satisfying this condition is \( n = 9 \). Thus, controlling the remainder by the 9th term ensures that the partial sum is within \( 0.001 \) of the series sum.

Key concepts

Convergent Alternating Series

A convergent alternating series is a series that has terms alternating in sign (positive, negative, and so on) and converges to a finite limit. This means the series approaches a specific number as you add an infinite number of terms. An alternating series can be represented as:\[ \sum_{n=1}^{\infty} (-1)^{n+1} a_n \]Here, each term of the series alternates in sign due to \((-1)^{n+1}\), and the sequence of term magnitudes \(a_n\) usually decreases as \(n\) increases.To be classified as a convergent alternating series, it should meet certain conditions:

• The absolute values of the terms, \(a_n\), should decrease to zero as \(n\) increases, i.e., \(a_n \to 0\).
• The sum of all terms should either converge to a finite number or approach zero as more terms are included.

Understanding this concept is crucial for applying the Alternating Series Remainder Theorem, which helps determine how close partial sums get to the actual sum of the series.

Partial Sum

In a series, a partial sum represents the sum of the first few terms. For our alternating series, the \(n^{\text{th}}\) partial sum \(S_n\) is given by adding up the series terms up to \(n\):\[ S_n = \sum_{k=1}^{n} (-1)^{k+1} a_k \]Partial sums are useful because they help estimate how much of the series sum we've accounted for after a fixed number of terms. By calculating \(S_n\), we can get a rough idea of what the overall series sum might be.In the context of a convergent alternating series, as the number of terms (\(n\)) included in the partial sum increases, \(S_n\) gets closer to the actual sum \(S\) of the entire series. Partial sums give us a practical way to approximate the sum without needing to compute an infinite number of terms.

Inequality

When tasked with ensuring that the partial sum is within a certain error margin \(\varepsilon\) from the series sum \(S\), inequalities play a key role. The Alternating Series Remainder Theorem comes in handy here as it gives us a way to control the error.According to the theorem, the remainder \(R_n\)—which is the difference between the actual series sum \(S\) and the \(n^{\text{th}}\) partial sum \(S_n\)—can be estimated with the inequality:\[ |R_n| = |S - S_n| \leq a_{n+1} \]This tells us that the absolute value of the remainder is less than or equal to the absolute value of the following term \(a_{n+1}\). By solving the inequality \(a_{n+1} \leq \varepsilon\), we keep the partial sum within the desired accuracy range \(\varepsilon\).For example, if \(\varepsilon = 0.001\), we solve for \(n\) such that \(a_{n+1} = \frac{1}{(n+1)^4}\) is less than or equal to \(0.001\). Solving this leads to determining the minimum \(n\) that satisfies this, ensuring that our partial sum is suitably close to the actual sum.