Is series $\sum_{n=0}^{\infty} \Big(\frac{x}{1-x}\Big)^n \frac1{(1-x)^2}$ uniformly convergent for $x <\frac1{2}$?

Question

I believe by the ratio test that it converges for $x < \frac1{2}$ but I can't seem to apply a Weierstrauss M test or other test to show uniform convergence. Maybe it is not.

Answer 1

No. It's not uniformly convergent for $x<1/2$.

Let $$f_N(x) = \sum_{n=0}^{N} \Big(\frac{x}{1-x}\Big)^n \frac1{(1-x)^2}$$ and $$f(x) = \sum_{n=0}^{\infty} \Big(\frac{x}{1-x}\Big)^n \frac1{(1-x)^2}$$ for $x<1/2$.

If $f_N$ was uniformly convergent to $f$ for $x<1/2$, then you'd be able to find $N$ such that $| f - f_N |$ is bounded by $1$ for $x<1/2$. But $f(x) - f_N(x) = \frac{1}{(1-x)} \frac{1}{(1-2x)} \Big( \frac{x}{1-x}\Big)^{N+1} $ is clearly unbouded for $x<1/2$.