Determine how many terms of the following convergent series must be summed to be sure that the remainder is less than 

10 Superscript negative 610−6 

in magnitude. 

Summation from k equals 0 to infinity StartFraction left parenthesis negative 1 right parenthesis Superscript k Over left parenthesis 2 k plus 3 right parenthesis Superscript 4 EndFraction∑k=0∞(−1)k(2k+3)4

Question content area bottom

Part 1

The number of terms that must be summed is 

enter your response here.

​(Round up to the nearest integer as​ needed.) 

To solve this problem, we’ll analyze the alternating series remainder to determine how many terms of the given series must be summed to ensure that the remainder is less than \( 10^{-6} \) in magnitude.

### Step 1: Understanding the Series and Alternating Series Remainder Theorem

The series is given by:
\[
\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+3)^4}.
\]

Since it is an alternating series with terms of the form \(\frac{(-1)^k}{(2k+3)^4}\), we can use the **Alternating Series Remainder Theorem**. This theorem states that for a convergent alternating series, the remainder \( R_n \) after summing \( n \) terms satisfies:
\[
|R_n| \leq |a_{n+1}|,
\]
where \( a_{n+1} \) is the \((n+1)\)-th term of the series. Therefore, to ensure that the remainder is less than \( 10^{-6} \), we need:
\[
\left|\frac{1}{(2(n+1)+3)^4}\right| < 10^{-6}.
\]

### Step 2: Setting Up the Inequality

The general term of the series is \( \frac{1}{(2k+3)^4} \). For the \((n+1)\)-th term, this becomes:
\[
\frac{1}{(2(n+1) + 3)^4}.
\]
We need this term to be less than \( 10^{-6} \):
\[
\frac{1}{(2(n+1) + 3)^4} < 10^{-6}.
\]

### Step 3: Solving for \( n \)

1. Invert both sides of the inequality:
   \[
   (2(n+1) + 3)^4 > 10^6.
   \]

2. Take the fourth root of both sides:
   \[
   2(n+1) + 3 > 100.
   \]

3. Solve for \( n \):
   \[
   2(n+1) > 97,
   \]
   \[
   n+1 > 48.5,
   \]
   \[
   n > 47.5.
   \]

Since \( n \) must be an integer, we round up to ensure the remainder is below the required tolerance. Thus, we find:
\[
n = 48.
\]

### Conclusion

To ensure the remainder is less than \( 10^{-6} \) in magnitude, **we must sum the first 48 terms** of the series.

---

Would you like further details or have any questions? 

Here are 5 related questions to explore further:

1. How does the Alternating Series Remainder Theorem apply to non-alternating series?
2. Can we estimate the convergence speed for this type of series?
3. What is the practical application of ensuring the remainder is within a specific tolerance?
4. How does this series compare to other convergent series with similar forms?
5. What changes in the solution if the tolerance were different, e.g., \( 10^{-5} \)?

**Tip:** For convergent alternating series, always check if the absolute value of each term decreases as it progresses, as this confirms the use of the Alternating Series Remainder Theorem.

Determine how many terms of the following convergent series must be summed to be sure that the remainder is less than 10^(-6) in magnitude. 
Sum from k = 0 to infinity of (-1)^k / (2k + 3)^4.
The number of terms that must be summed is (Round up to the nearest integer as needed).

Learn how many terms of the convergent series Sum from k = 0 to infinity of (-1)^k / (2k + 3)^4 are required to ensure the remainder is below 10^-6. This solution uses the Alternating Series Remainder Theorem and provides a step-by-step guide. Ideal for students studying calculus and series convergence.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation