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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?

Olba12
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  • You have to be able to do something like $\frac{b-a}{N/2}\sum f(a+k\frac{b-a}{N/2})$, $f$ being the integrand. Try it, it's just a matter of "correcting" the faulty $k$ under the sinus. – Nicolas FRANCOIS Nov 15 '16 at 00:54
  • I have mad an edit @NicolasFRANCOIS – Olba12 Nov 15 '16 at 01:06
  • I dont remember the rules for when the sum is equal to an integral, and I dont now how to continue now. @NicolasFRANCOIS – Olba12 Nov 15 '16 at 01:12
  • I meant $N/2$ because you have half the $k$ (only the odd ones). Your $\Delta x$ should be $\frac{k\pi}{N+1}$. Sorry, don't have time for more advice now, I have to go to work :-) – Nicolas FRANCOIS Nov 15 '16 at 07:48

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