Approximate the sum of the series with error less than 0.0001. In other words, find sn for the value of n found in part a. Round your answer to 4 decimal places. 1/((n)(7^n)) n = 4

To approximate the sum of the series \(\sum \frac{1}{n \cdot 7^n}\) with an error less than \(0.0001\), we need to calculate the partial sum \(S_n\) up to a certain value of \(n\), such that the remainder term \(R_n\) satisfies:

\[
R_n = \sum_{k=n+1}^{\infty} \frac{1}{k \cdot 7^k} < 0.0001
\]

Let's go through the steps:

1. **Calculate the Partial Sum**: We calculate the partial sum up to \(n = 4\) as requested:
   \[
   S_4 = \frac{1}{1 \cdot 7^1} + \frac{1}{2 \cdot 7^2} + \frac{1}{3 \cdot 7^3} + \frac{1}{4 \cdot 7^4}
   \]

2. **Compute Each Term in \(S_4\)**:
   - \( \frac{1}{1 \cdot 7^1} = \frac{1}{7} \approx 0.1429 \)
   - \( \frac{1}{2 \cdot 7^2} = \frac{1}{98} \approx 0.0102 \)
   - \( \frac{1}{3 \cdot 7^3} = \frac{1}{1029} \approx 0.0010 \)
   - \( \frac{1}{4 \cdot 7^4} = \frac{1}{9604} \approx 0.0001 \)

3. **Add the Terms**:
   \[
   S_4 \approx 0.1429 + 0.0102 + 0.0010 + 0.0001 = 0.1542
   \]

Thus, the approximate sum \(S_4\) is:

\[
\boxed{0.1542}
\]

Would you like to see the error estimation or further explanation of this solution?

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Here are five related questions:
1. How do we estimate the error in a convergent series?
2. Why does the error decrease with each additional term in a convergent series?
3. What are some common series that converge faster than this one?
4. How does the base \(7\) in the denominator affect the rate of convergence?
5. How would the series change if we used a different base instead of \(7\)?

**Tip:** When working with series, always check if there’s a remainder or error term approximation to ensure accuracy within your desired tolerance.

Learn how to approximate the sum of the series 1/(n*7^n) for n=4 with an error tolerance below 0.0001. This solution includes calculating partial sums, using error approximation, and understanding series convergence. Ideal for high school students in Grades 10-12.

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