# Uniform convergence of the sequence $\frac{\ln(1+\frac xn)}{(x+1)}$

How do I show that $\displaystyle f_n(x) = \frac{\ln(1 +\frac xn)}{x+1}$ converges uniformly on $\displaystyle [0,a]$ for $\displaystyle a > 0$?

Clearly the sequence converges pointwise to $\displaystyle 0$, but I don't know where to start to prove uniform convergence?

Also, is the sequence uniformly convergent on $\displaystyle [0, \infty)$?

• I think you're mistaken. The sequence doesn't converge pointwise to $0$ at all, only at $x = 1$, and it definitely doesn't converge uniformly on any $[0, a]$; it's not even defined at $x = 0$. Commented Jun 6, 2018 at 19:01
• Sorry, I have mistakenly written x instead of 1, it's edited now.
– Dark
Commented Jun 6, 2018 at 19:05
• $f_n = \frac {\ln(1+ \frac {x}{n})^n}{n(x+1)}\\\lim_\limits{n\to \infty} (1+\frac {x}{n})^n = e^x$ Commented Jun 6, 2018 at 19:16

Note that we have for $x\ge 0$
$$\left|\frac{\log\left(1+\frac xn\right)}{x+1}\right|\le \frac{x}{n(1+x)}<\frac1n$$
which proves uniform convergence for $x\ge 0$