Show that the sequence $\left\{\frac{1}{1+nz}\right\}$ is uniformly convergent to zero for all $z$ such that $|z|\geq 2$.
The definition for uniform convergence is:
We say $\left\{f_n\right\}$ converges to $f$ uniformly on a subset $S$ of $\Omega$ iff for every $\epsilon >0$ there is an $n_0\in\mathbb{N}$ so that if $n>n_0$ then $|f_n(z)-f(z)|<\epsilon$ for all $z\in S$.
I found this example in my book, but I don't understand how they proved it with definition. All it says is:
$|\frac{1}{1+nz}|\leq \frac{1}{n|z|-1}\leq\frac{1}{2n-1}\leq\frac{1}{n}$.
Best Answer
They have used reverse triangle inequality to derive that inequality.
Note that: to show that $f_n \to f$ uniformly on $S$, it suffices to show that $$\sup_{z\in S} |f_n(z) - f(z)| \to 0.$$ (Why? Use the definition given to you as well as the definition of the limit of a sequence of real numbers.)
In this case, you have that $|f_n(z) - 0| \le 1/n$ for all $z \in S$ and hence, $$\sup_{z\in S} |f_n(z) - f(z)| \le \dfrac{1}{n}.$$ Using the Sandwich theorem, we see that $\displaystyle\sup_{z\in S} |f_n(z) - f(z)| \to 0$, as desired.