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.