Show that $$\lim_{N \to \infty} \frac{4}{\pi} \sum_{ k > 0, \ k \text{ odd }}^{N} \frac{\sin({\frac{k\pi}{N+1}})}{k} = \frac{2}{\pi}\int_{0}^{\pi} \frac{\sin (x)}{x} dx$$
I thought that I write the left hand side as a Riemann sum.
$$\frac{4}{\pi} \sum_{ k > 0, \ k \text{ odd }}^{N} \frac{\sin({\frac{k\pi}{N+1}})}{k} = \frac{4}{\pi} \sum_{ k > 0, \ k \text{ odd }}^{N} \frac{\sin({\frac{k\pi}{N+1}})}{\frac{2k \pi}{N+1}} \frac{2\pi}{N+1} = \frac{2}{\pi} \sum_{ k > 0, \ k \text{ odd }}^{N} \frac{\sin(k \frac{\Delta x}{2})}{k \frac{\Delta x}{2}} \Delta x $$
And then this is the middle Riemann sum, which converges to the integral. Correct?