Check the uniform convergence of the sequence of functions $f_n(x)=n\ln \left(\frac{1+nx}{nx}\right)$ on $(0,\infty)$.
I have found that the pointwise limit of the sequence of functions is $f(x)=\frac 1x$. I'm stuck proving uniform convergence. I am unable to find $\displaystyle M_n=\sup_{x\in (0,\infty)}\left|n\ln \left(\frac{1+nx}{nx}\right)-\frac 1x\right|$.
Any hints, please?
The convergence is not uniform on any interval of the form $(0,a)$. Note that
$$n\ln \left(\frac{1+nx}{nx}\right)-\frac 1x = n\ln \left(\frac{1+nx}{nx}\right)-n\ln e^{\frac{1}{nx}} = n\ln \left(\frac{1+nx}{nxe^{\frac{1}{nx}} }\right),$$
and with $x = 1/n \in (0,1)$,
$$\sup_{x\in (0,\infty)}\left|n\ln \left(\frac{1+nx}{nx}\right)-\frac 1x\right| \geqslant \left|n\ln \left(\frac{1+n\cdot \frac{1}{n}}{n\cdot \frac{1}{n}\cdot e^{\frac{1}{n\cdot \frac{1}{n}}} }\right)\right|= n \ln \frac{e}{2} \underset{n \to \infty}\longrightarrow \infty$$