The sequence $n\sin(\sqrt{4\pi^2n^2 +x^2})$ converges on compacts:

Question

I'm not sure about the idea behind the sequence $f_n(x)=n\sin(\sqrt{4\pi^2n^2 +x^2})$ being uniformly convergent on every compact of the form $[0,a]$ for $a>0$. On $\mathbb{R}$ we can find a sequence such that $\sin$ alternates the sign, thus making it impossible for $f_n$ to have a limit.

But the compact case is a little more tricky: let $K=[0,a]$, then by Weierstrass Thm $\forall n\exists M_n\mid$ $|f_n(x)|<M_n$, thus we wish for $M_n$ to be convergent. As we are using Weierstrass Thm, we can take $x_n\in K$ and $f_n(x_n)=M_n$. But $K$ is compact in $\mathbb{R}$: $\exists x_{n_k}\mid x_{n_k}\rightarrow\bar{x}\in K$, thus $f_{n_k}(x_{n_k})=M_{n_k}\rightarrow\overline{M}\in\mathbb{R}$, making $f_n$ uniformly convergent. The proof seems to make sense but on the other hand it doesn't depend on $f_n$, so it sounds like I'm either missing something.

Answer 1

HINT:

Note that we have for any fixed $x$

$$\begin{align} \sin(\sqrt{4\pi n^2+x^2})&=\sin\left(2\pi n\sqrt{1+\left(\frac{x}{2\pi n}\right)^2}\right)\\\\ &=\sin\left(2\pi n \left(1+\frac{x^2}{8\pi n^2}+O(n^{-4})\right)\right)\\\\ &=\frac{x^2}{4\pi n}+O\left(n^{-3}\right) \end{align}$$

whence multiplying by $n$ and letting $n\to \infty$ yields

$$\lim_{n\to \infty}n\sin(\sqrt{4\pi n^2+x^2})=\frac{x^2}{4\pi}$$

Answer 2

What you are missing is to actually use the functions you have.

Concretely, we have $$ f_n(x)=n\,\sin\Big(2\pi n\,\sqrt{1+\frac{x^2}{4\pi^2n^2}}\Big). $$ Using the binomial expansion, $$ \sqrt{1+\frac{x^2}{4\pi^2n^2}}=1+\frac{x^2}{8\pi^2n^2}+o(n^{-4}). $$ The sine is $$ \sin t= t - o(t^3). $$ Then \begin{align} f_n(x)&=n\,\sin\Big(2\pi n+\frac{2\pi nx^2}{8\pi^2n^2}+o(n^{-3})\Big)\\[0.3cm] &=n\,\sin\Big(\frac{2\pi nx^2}{8\pi^2n^2}+o(n^{-3})\Big)\\[0.3cm] &=\frac{2\pi n^2x^2}{8\pi^2n^2}-o(n^{-2})\\[0.3cm] &=\frac{ x^2}{4\pi}-o(n^{-2})\\[0.3cm] &\xrightarrow[n\to\infty]{}\frac{x^2}{4\pi} \end{align}

When we consider $x$ only on some compact set, because it's bounded the convergence can be made uniform, as the estimates will not depend on $x$ (hidden, above in the terms $o(n^{-4})$, etc.)