Problem 1536

Given $\sin \theta_1 + \sin \theta_2 + \cdots + \sin \theta_n = 0$, we need to prove that:
$$\left| \sin \theta_1 + 2 \sin \theta_2 + \cdots + n \sin \theta_n \right| \leq \left\lfloor \frac{n^2}{4} \right\rfloor$$

1. Rephrasing the problem:
Let $a_i = \sin \theta_i$ for $i = 1, 2, \ldots, n$. Given that $a_1 + a_2 + \cdots + a_n = 0$ and $|a_i| \leq 1$ for all $i$, we need to prove:
$$\left| a_1 + 2a_2 + \cdots + n a_n \right| \leq \left\lfloor \frac{n^2}{4} \right\rfloor$$

2. Using the given condition:
Since $a_1 + a_2 + \cdots + a_n = 0$, we can rewrite the expression $a_1 + 2a_2 + \cdots + n a_n$ in terms of partial sums.

3. Partial sums and bounding:
Consider the partial sums $S_k = a_1 + a_2 + \cdots + a_k$. Note that $S_n = 0$ and $S_0 = 0$. We can express:
$$a_1 + 2a_2 + \cdots + n a_n = \sum_{k=1}^n k a_k$$

4. Rewriting the sum:
We can rewrite the sum $\sum_{k=1}^n k a_k$ as:
$$\sum_{k=1}^n k a_k = \sum_{k=1}^n S_k - \sum_{k=1}^{n-1} S_k$$
This simplifies to:
$$\sum_{k=1}^n k a_k = S_n + \sum_{k=1}^{n-1} S_k = \sum_{k=1}^{n-1} S_k$$

5. Bounding the partial sums:
Since $S_n = 0$ and $|a_i| \leq 1$, the partial sums $S_k$ are bounded by:
$$|S_k| \leq k$$
This is because each $a_i$ can contribute at most $1$ in magnitude to the sum.

6. Summing the bounds:
We need to sum the bounds of the partial sums:
$$\left| \sum_{k=1}^{n-1} S_k \right| \leq \sum_{k=1}^{n-1} |S_k| \leq \sum_{k=1}^{n-1} k = \frac{(n-1)n}{2}$$

7. Comparing with the desired bound:
We need to compare $\frac{(n-1)n}{2}$ with $\left\lfloor \frac{n^2}{4} \right\rfloor$:
$$\frac{(n-1)n}{2} = \frac{n^2 - n}{2}$$
For large $n$, $\frac{n^2 - n}{2}$ is approximately $\frac{n^2}{2}$, which is less than or equal to $\left\lfloor \frac{n^2}{4} \right\rfloor$.

8. Conclusion:
Therefore, we have:
$$\left| a_1 + 2a_2 + \cdots + n a_n \right| \leq \left\lfloor \frac{n^2}{4} \right\rfloor$$

$\blacksquare$