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I can't come up with a proof, why $f_N(y) := \frac{\frac{y}{2N}}{\sin\left(\frac{y}{2N}\right)}$ converges uniformly against $1$ for $y\in(0,\pi),\ N\to\infty$.

I would be thankful for any advice.

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    $\begingroup$ One idea is to use $x-x^3/6 \leq \sin(x) \leq x$ which holds for $x \geq 0$. $\endgroup$ – Ian Jan 4 at 15:15
  • $\begingroup$ Ian.Want to post an answer with your idea? $\endgroup$ – Peter Szilas Jan 4 at 15:32
  • $\begingroup$ What do you mean by $y/2N$? $\frac{y}{2N}$ or $\frac{y}{2}N$? $\endgroup$ – mathcounterexamples.net Jan 4 at 15:58
  • $\begingroup$ @mathcounterexamples.net For the result to be as they say it must be the former... $\endgroup$ – Ian Jan 4 at 16:02
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    $\begingroup$ @stressedout The poles are nowhere to be seen here, even when $N=1$, because of the $2$. $1-x^2/6 \leq \sin(x)/x \leq 1$ simply becomes $1 \leq x/\sin(x) \leq \frac{1}{1-x^2/6}$ which is valid for $0<x<\sqrt{6}$, and $\pi/2<\sqrt{6}$. $\endgroup$ – Ian Jan 4 at 16:10
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A solution using Dini's theorem

The $f_n$ can be extended by continuity at $0$ by raising $f_n(0)=1$. Hence, we can consider the continuous extended maps on the compact interval $[0,\pi]$.

For $n \ge 2$ the sequence $(f_n)$ is uniformly bounded below by a constant strictly positive. Hence proving the uniform convergence of $(f_n)$ is equivalent to the proof of the uniform convergence of $(1/f_n)$. And this is provided by Dini's theorem as for all $x \in [0, \pi]$ the sequence $(1/f_n(x))$ is increasing.

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