# Uniform convergence of $\frac{y/(2N)}{\sin(y/(2N))}$ towards 1

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.

• One idea is to use $x-x^3/6 \leq \sin(x) \leq x$ which holds for $x \geq 0$. – Ian Jan 4 at 15:15
• Ian.Want to post an answer with your idea? – Peter Szilas Jan 4 at 15:32
• What do you mean by $y/2N$? $\frac{y}{2N}$ or $\frac{y}{2}N$? – mathcounterexamples.net Jan 4 at 15:58
• @mathcounterexamples.net For the result to be as they say it must be the former... – Ian Jan 4 at 16:02
• @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 $\pi/2<\sqrt{6}$. – Ian Jan 4 at 16:10

## 1 Answer

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.