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.