There is another way of doing this computation. You correctly pointed out that: $$\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$$
so:
$$\frac{1}{(1-x)^2} = \frac{1}{1-x}\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n\sum_{n=0}^{\infty} x^n = \sum_{n=0}^{\infty}\sum_{k=0}^{n}x^kx^{n-k} = \sum_{n=0}^{\infty}(n+1)x^n$$
This series is convergent since it is the Cauchy product of two absolutely convergent series.
Now we can complete using the same reasoning you used:
$$4x^3\dfrac{1}{(1-x)^2} = 4x^3\sum_{n=0}^{\infty}(n+1)x^n = \sum_{n=0}^{\infty}4(n+1)x^{n+3}$$
To answer your question in the comments, you cannot easily exchange differentiation and summation since you require uniform convergence of the derivatives, which in this case you don't have.
Edit:
Since you haven't seen the Cauchy product we can do it with the monotone convergence theorem. Call $f_m(x) = \sum_{n=0}^{m}(n+1)x^n$ and $f(x) = \sum_{n=0}^{\infty}(n+1)x^n$, then $$\int f_m(x)dx = \int \sum_{n=0}^{m}(n+1)x^n dx = \sum_{n=0}^{m}\int(n+1)x^n dx$$ since this time the sum is finite and we have no problem integrating term by term. Now if we manage to prove $$\sum_{n=0}^{\infty}\int(n+1)x^n dx = \lim_{m\to\infty} \int f_m(x) dx = \int f(x) dx = \int \sum_{n=0}^{\infty}(n+1)x^n dx $$ we will have that $$\frac{d}{dx}\sum_{n=0}^{\infty} x^n = \sum_{n=0}^{\infty}(n+1)x^n$$ and then we will be able to proceed as before.
To do this we apply the dominated convergence theorem, and since clearly $f_m(x) > 0$ for all $x \in (-1, 1)$ we just need to show that $f \geq f_m$:
$$f - f_m = \sum_{n=0}^{\infty}(n+1)x^n - \sum_{n=0}^{m}(n+1)x^n = \sum_{n=m}^{\infty}(n+1)x^n$$
which is just the original series up to rearrangement, so it is also greater than $0$ and we are done.
