Why does $\forall x \in \mathbb{R}$, $ \forall n \in \mathbb{N}$ :$|\text{sin}(\frac{x}{n^2})| \le \frac{|x|}{n^2} $?

Question

I'm studying the convergence and absolute convergence of the series of functions defined by the sequence of functions: \begin{equation*} f_n: \mathbb{R} \to \mathbb{R}, \end{equation*}

\begin{equation*} \phantom{1000}x \mapsto \sin\left(\dfrac{x}{n^2}\right). \end{equation*}

If I get that $\forall x \in \mathbb{R}$, $\forall n \in \mathbb{N}$: \begin{equation*} \left|\sin\left(\dfrac{x}{n^2}\right)\right| \le \dfrac{|x|}{n^2}, \end{equation*} I could apply the comparison criteria for series

Answer 1

You know $|\sin y|\le |y|$ for all real $y$ hence $ \left|\sin\left(\frac{x}{n^2}\right)\right| \le \frac{|x|}{n^2}. $

For this just observe that $$g(x)=x-\sin x$$ is a increasing since $g'(x)\ge0$ function then you will have $g(x)> 0$ for $x> 0$ and $g(0)=0$.

So yes, you are right but this comparison would not help you to show the uniform convergence . If you had the domain of $f_n$'s to be a bounded interval ,say $[-M,M]$ then you had this nice inequality $$ \left|\sin\left(\dfrac{x}{n^2}\right)\right| \le \dfrac{|x|}{n^2}\le\dfrac{M}{n^2},\forall x \in [-M,M] \text{ and }\forall n \in \mathbb{N} $$
Then by comparison test and Weierstrass M-test you have the sequence of function $\{f_n\}$ as well as the series $\sum f_n $ converges uniformly on $[-M,M]$.

But on $\mathbb R$ the sequence converges point-wise to $0$,that is $$ \lim_{n\to \infty}f_n(x)=\lim_{n\to \infty}\sin\left(\frac{x}{n^2}\right)=0,\forall x\in \mathbb R. $$ But the sequence doesn't converge uniformly.

Since $$f_n\left(\frac{n^2\pi}{2}\right)=\sin \left(\frac{\pi}{2}\right)=1,\forall n\in\mathbb N. $$

Answer 2

You have that $\left|\sin\left(\dfrac{x}{n^2}\right)\right| \le \dfrac{|x|}{n^2}$ because

$$ \left|\sin\left(y \right)\right| \le |y| $$

for every real $y$.

To prove this you just have to consider the case $0<x<1$, because you already know that $\sin y$ cannot be bigger than $1$. In $0<x<1$ the derivative of $\sin y$ is $\cos y$, which is positive and smaller than $1$: the function $\sin y$ is increasing slower than $y$ in this interval. Therefore, since both $\sin y$ and $y$ are both zero at the origin, you can conclude that $y>\sin y$ over $0<x<1$, and then also over the positive real line. Now you only have to apply the modulus and extend the result to the whole real line.