103 Trigonometry Problems

5. Solutions to Advanced Problems 197
Here we set the first summation on the right-hand side to be 0 if m = 0, and
the first summation taken from 1 to n and the second to be 0 if m ≥ n. It
suffices to show that ∣ ∣∣∣∣ ∑
m sin kx
k ∣ ≤ √ π
(∗)
k=1
and ∣ ∣∣∣∣ n∑
k=m+1
sin kx
k
∣ ≤ √ π.
Because | sin x|