The sequence $(s_n)$, where $s_n= \sum_{k=1}^n \sin k$ for $n = 1, 2,\dots$ is bounded. Find an upper and lower bound.
1. **Express the sum in terms of complex exponentials:**
   We start by using Euler's formula, which states that $\sin k = \frac{e^{ik} - e^{-ik}}{2i}$. Therefore, the sum $s_n$ can be written as:
   \[
   s_n = \sum_{k=1}^n \sin k = \sum_{k=1}^n \frac{e^{ik} - e^{-ik}}{2i}
   \]
   This can be split into two separate sums:
   \[
   s_n = \frac{1}{2i} \left( \sum_{k=1}^n e^{ik} - \sum_{k=1}^n e^{-ik} \right)
   \]

2. **Evaluate the geometric series:**
   Each of these sums is a geometric series. Recall that the sum of a geometric series $\sum_{k=0}^{n-1} ar^k$ is given by $a \frac{r^n - 1}{r - 1}$. Applying this to our sums:
   \[
   \sum_{k=1}^n e^{ik} = e^i \frac{e^{in} - 1}{e^i - 1} = \frac{e^{i(n+1)} - e^i}{e^i - 1}
   \]
   and
   \[
   \sum_{k=1}^n e^{-ik} = e^{-i} \frac{e^{-in} - 1}{e^{-i} - 1} = \frac{e^{-i(n+1)} - e^{-i}}{e^{-i} - 1}
   \]

3. **Simplify the sums:**
   Simplifying these expressions, we get:
   \[
   \sum_{k=1}^n e^{ik} = \frac{e^{i(n+1)} - e^i}{e^i - 1} = \frac{e^{i(n+1)} - e^i}{e^{i} (e^{i(n+1)} - 1)}
   \]
   and
   \[
   \sum_{k=1}^n e^{-ik} = \frac{e^{-i(n+1)} - e^{-i}}{e^{-i} - 1} = \frac{e^{-i(n+1)} - e^{-i}}{e^{-i} (e^{-i(n+1)} - 1)}
   \]

4. **Combine the results:**
   Combining these results, we have:
   \[
   s_n = \frac{1}{2i} \left( \frac{e^{i(n+1)} - e^i}{e^i - 1} - \frac{e^{-i(n+1)} - e^{-i}}{e^{-i} - 1} \right)
   \]

5. **Simplify further:**
   Simplifying the above expression, we get:
   \[
   s_n = \frac{1}{2i} \left( \frac{e^{i(n+1)} - e^i}{e^i - 1} - \frac{e^{-i(n+1)} - e^{-i}}{e^{-i} - 1} \right)
   \]
   This can be further simplified to:
   \[
   s_n = \frac{1}{2i} \left( \frac{e^{i(n+1)} - e^i}{e^i - 1} - \frac{e^{-i(n+1)} - e^{-i}}{e^{-i} - 1} \right)
   \]

6. **Use trigonometric identities:**
   Using the identity $\cos x = \frac{e^{ix} + e^{-ix}}{2}$ and $\sin x = \frac{e^{ix} - e^{-ix}}{2i}$, we can rewrite the sum as:
   \[
   s_n = \frac{1}{2i} \left( \frac{e^{i(n+1)} - e^i}{e^i - 1} - \frac{e^{-i(n+1)} - e^{-i}}{e^{-i} - 1} \right)
   \]

7. **Find the bounds:**
   The bounds of the sequence can be found by considering the maximum and minimum values of the cosine function. Since $\cos x$ is bounded between -1 and 1, the sequence $s_n$ is also bounded. Specifically, the bounds are:
   \[
   -\frac{1}{2} \leq s_n \leq \frac{1}{2}
   \]

The final answer is $\boxed{-\frac{1}{2} \leq s_n \leq \frac{1}{2}}$