Answer
See explanation
Work Step by Step
**Solution Explanation**
We have a sequence defined by the formula
\[
s_n = \frac{(-1)^n}{n!}, \quad \text{for all integers } n \ge 0.
\]
We want to show that it satisfies the recurrence
\[
s_k = \;-\;\frac{s_{k-1}}{k} \quad \text{for all integers } k \ge 1.
\]
---
### Step 1. Write Down \(s_k\) and \(s_{k-1}\)
From the given explicit formula:
\(s_k = \dfrac{(-1)^k}{k!}\).
\(s_{k-1} = \dfrac{(-1)^{k-1}}{(k-1)!}\).
---
### Step 2. Verify the Recurrence
Compute \(-\dfrac{s_{k-1}}{k}\):
\[
-\;\frac{s_{k-1}}{k}
\;=\;
-\;\frac{\displaystyle \frac{(-1)^{k-1}}{(k-1)!}}{k}
\;=\;
-\;\frac{(-1)^{k-1}}{k \,(k-1)!}
\;=\;
-\;\frac{(-1)^{k-1}}{k!}.
\]
Since \(-\,(-1)^{k-1} = (-1)^k,\) we get
\[
-\;\frac{(-1)^{k-1}}{k!}
\;=\;
\frac{(-1)^k}{k!}
\;=\;
s_k.
\]
Hence,
\[
s_k = \;-\;\frac{s_{k-1}}{k},
\]
which completes the proof.
