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Let's solve part **d** of the given problem, which involves proving that the infinite series converges to 5:

We are asked to show that:

\[
s = \lim_{n \to \infty} s_n = \sum_{k=1}^{\infty} \frac{4}{5^{k-1}} = 5
\]

### Step-by-Step Solution:

1. **Recognize the series**: The given series is a geometric series. A general geometric series is of the form:

   \[
   \sum_{k=0}^{\infty} ar^k
   \]
   where \(a\) is the first term and \(r\) is the common ratio.

2. **Rewrite the series**: We can rewrite the given series in a more familiar form. Notice that the first term corresponds to \(k = 1\), but for simplicity, we can change the index by letting \(k' = k - 1\). So:

   \[
   \sum_{k=1}^{\infty} \frac{4}{5^{k-1}} = \sum_{k'=0}^{\infty} \frac{4}{5^{k'}}
   \]
   This is now a geometric series with \(a = 4\) and \(r = \frac{1}{5}\).

3. **Sum of an infinite geometric series**: The sum of an infinite geometric series is given by:

   \[
   S = \frac{a}{1 - r}
   \]
   provided that \(|r| < 1\), which is true in our case since \(r = \frac{1}{5}\).

4. **Apply the formula**:
   
   Here, \(a = 4\) and \(r = \frac{1}{5}\). So:

   \[
   S = \frac{4}{1 - \frac{1}{5}} = \frac{4}{\frac{4}{5}} = 5
   \]

Thus, we have shown that the sum of the infinite series is 5, which completes the proof.

---

Would you like more details or have any further questions?

### Follow-up questions:
1. Can you explain why the condition \(|r| < 1\) is necessary for the convergence of a geometric series?
2. What would happen if \(r \geq 1\) in a geometric series?
3. Could you apply the same method to a series with a different first term or ratio? How?
4. How does changing the index from \(k\) to \(k-1\) affect the structure of the series?
5. What would the sum be if the first term \(a\) were different, say \(a = 6\)?

### Tip:
For a geometric series, always ensure that the common ratio \(r\) is between \(-1\) and \(1\) to guarantee convergence.

Aufgabe 3.9: a_n = 4 / (5^n), n ∈ N. d) Zeigen Sie allgemein, dass für die Reihe s = lim(n → ∞) s_n gilt: s = Σ(k=1,∞) (4 / (5^(k-1))) = 5.

Solve the geometric series problem where a_n = 4 / (5^n) and show that the series converges to 5. Learn about limits, infinite geometric series, and their sum formulas. This is suitable for undergraduate students or advanced high school learners.

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