Find the range where $f_n = \frac{1}{ 1- x^n}$ is uniformly convergent.

Question

I have tried dividing the interval into |x| >1 and |x|<1 and using wierstrass M- test but failed. I am not sure how to approach the problem. How do I start the problem?

Answer 1

Note that we have pointwise

$$\lim_{n\to \infty}\frac{1}{1-x^n}=\begin{cases} 1&,|x|<1\\\\ 0&,|x|>1 \end{cases}$$

We shall show now that the sequence converges uniformly for $|x|\le r<1$ for any $r<1$ and for $|x|\ge r>1$ for any $r>1$.

Let $\epsilon>0$ be given. For $|x|\le r<1$ we have

$$\left|\frac{1}{1-x^n}-1\right|\le \frac{|r|^n}{1-|r|^n}<\epsilon$$

whenever $n>\frac{\log\left(\frac{\epsilon}{1+\epsilon}\right)}{\log(|r|)}$.

For $|x|\ge r>1$ we have

$$\left|\frac{1}{1-x^n}-0\right|\le \frac{1}{|r|^n-1}<\epsilon$$

whenever $n>\frac{\log\left(\frac{1+\epsilon}{\epsilon}\right)}{\log(|r|)}$.