Originally Posted by

**hmmmm** The function $\displaystyle f_n=\frac{x^n}{x^n+1}$ converges uniformly to $\displaystyle f(x)=1$ on the interval $\displaystyle (1,a)$.

Proof.

taking x=a

as $\displaystyle |\frac{a^n}{a^n+1}-1|<\epsilon\rightarrow \frac{1}{1+a^n}<\epsilon \rightarrow ln(\frac{1-\epsilon}{\epsilon})<ln(a) n$

So can I take $\displaystyle N={ln(\frac{1-\epsilon}{\epsilon})}$ and then I have that:

$\displaystyle \forall\epsilon>0\exists N(={ln(\frac{1-\epsilon}{\epsilon})}) s.t \forall n\geq N \ \forall x\in(1,a)\ \mbox{we have}\ |f_n(x)-f(x)|<\epsilon$

is this correct or is it wildly off- uniform convergence is kind of killing me

thanks for any help