Prove that $(f_n)_{n\in\mathbb{N}},f_n:\mathbb{R}\rightarrow\mathbb{R}$ with $f_n(x)=\frac{n+1}{n}x$ is uniformly convergent.

Question

Maybe the Statement of the Question is also false, I am trying to come up with an easy example for uniform convergence.

It is clear that $\lim_{n\rightarrow\infty}(f_n)_{n\in\mathbb{N}}=$id,id$(x)=x$
To Show that the functionsequence is uniformly convergent I have to show

$\forall\epsilon>0\forall x\in \mathbb{R} \exists n_0\in\mathbb{N}\forall n \geq n_0:|\frac{n}{n+1}x-x|<\epsilon\iff |-\frac{1}{n+1}x|<\epsilon$

Now if I I Claim that I have found such a $n_0\in \mathbb{N}$ If I would pick an $\epsilon>1$ and $x=n_0+1$ I would get $1<1$

What am I doing wrong here?

Answer 1

Note that $|f_n(n)-f(n)| = 1$ for all $n$. Hence the convergence is not uniform.