Please check my work:
$$\frac{1}{1-x} = \sum_{k=0}^{\infty} x^k $$
for $|x|<1$ is the geometric series, which we know converges uniformly
Then by either differentiation or multiplication we see that
$$\frac{1}{(1-x)^2} = 1 + 2x + 3x^2 + ... = \sum_{k=1}^{\infty} k x^{k-1}$$
We check that $\sum_{k=1}^{\infty} k x^{k-1}$ converges with the Ratio Test
$$\lim_{k\to \infty} \left\lvert \frac{(k+1) x^k}{k x^{k-1}} \right\rvert =\lim_{k\to \infty} \left\lvert \frac{(k+1) x}{k} \right\rvert \le \lim_{k\to \infty} \left\lvert \frac{(k+1)}{k} \right\rvert \left\lvert x \right\rvert$$
$$\lim_{k\to \infty} \left\lvert \frac{(k+1)}{k} \right\rvert \left\lvert x \right\rvert = |x| <1 $$
So $\sum_{k=1}^{\infty} k x^{k-1}$ converges and since it is a Power Series in its interval of convergence, it converges absolutely and hence uniformly by the Weierstrauss M-test
Now since $\sum_{k=1}^{\infty} k x^{k-1}$ converges uniformly and the geometric series converges uniformly, it follows that
$$ \frac{1}{(1-x)^2} - \frac{1}{1-x}*\frac{1-x}{1-x} = \frac{x}{(1-x)^2} = \sum_{k=1}^{\infty} k x^{k-1} - \sum_{k=0}^{\infty} x^k = \sum_{k=1}^{\infty} kx^k $$
This proves that the function we are looking for is:
$$ \frac{x}{(1-x)^2} $$