# Determine whether the sequence is uniformly convergent.

Find out whether the following sequence is uniformly convergent on $$E$$. $$f_n(x)=\ln\left(\sin x+\frac{1}{n}\right),\ \ x\in E=\left(0;\frac{\pi}{6}\right)$$

Here is what I did: \left\{ \begin{aligned} &f(x)=\lim_{n\rightarrow\infty}\ln\left(\sin x+\frac{1}{n}\right)=\ln(\sin x)\\ &0<\sin x<\frac{1}{2} \end{aligned} \right.\Rightarrow \ln(\sin x)\ \exists\Rightarrow\\ \Rightarrow f_n(x) \text{ converges to } f(x)=\ln(\sin x)\\ t_n=|f_n(x)-f(x)|=\left|\ln\left(1+\frac{1}{n\sin x}\right)\right| Then I thought that $$\forall n\ \ \exists x: t_n > 0.\ \ \text{Thus, } \lim_{n\rightarrow\infty}\sup_{x\in E}t_n\ne0\Rightarrow f_n(x)\ \ \text{does not converge uniformly on E.}$$ Is my solution correct? And if so, please, tell me how to properly prove that $$\forall n\ \ \exists x:\ \lim_{n\rightarrow\infty}\sup_{x\in E}t_n\ne0.$$

Thank you in advance!

• If we take $x=\frac1{n^2}$ then $t_n$ is asymptotic with $\ln(1+n)$ which definitely doesn't go to $0$. – kingW3 Oct 11 '19 at 22:11

Note that $$\lim_{n \to \infty}n\, \sin \frac{1}{n} = 1$$.

Take $$x_n = \frac{1}{n} \in (0,\pi/6)$$ and find that as $$n \to \infty$$,

$$\sup_{x \in (0,\pi/6)}|f_n(x) - f(x) | \geqslant \left|\ln\left(1+\frac{1}{n\sin x_n}\right)\right| \to |\ln(2)| \neq 0$$