級数で書かれた関数の計算

  $\displaystyle f(x, n) = \frac{4}{\pi} \sum _{k=1}^{n} \frac{1}{2k-1} \sin (2k-1)x =\frac{4}{\pi} \left( \sin x + \frac{1}{3}\sin 3x +\frac{1}{5}\sin 5x + \cdots + \frac{1}{2n-1}\sin(2n-1)x \right) $