Let $ s =s(x)$ given by $s(x) = \sum_{k=1}^\infty (k+1)x^k. $ Prove that for all $ t \in ]-1,1[,$
$$\int_0^t s(x) dx = \frac{t^2}{1-t}$$
Conclude that, for all $x \in ]-1,1[,$
$$\sum_{k=1}^\infty (k+1)x^k = \frac{2x-x^2}{(1-x)^2}$$
My attempt:
To use the theorem about integration of series of functions, we must show, first, that the series is uniformly convergent to a function and continuous on a given interval.
I am not sure, but it looks like the series $\sum_{k=1}^\infty (k+1)x^k$ is not uniform convergent on $(-1,1)$ ($||Sn-Sm||=|\sum_{k=m+1}^n (k+1)|$), but as we can see (it is a power series), it is on every interval of the form $[-r,r], 0<r<1.$ For each $t \in (-1,1)$, we can find such $r$ so $ t \in [-r,r]$. Can I apply, now, the theorem about integration?
I am not sure if the same argument applies for the second part of the problem
Thanks in advance!
For $\;|t|<1\;$ we have
$$t^2\frac1{1-t}=t^2(1+t+t^2+\ldots+t^n+\ldots)=t^2+t^3+\ldots\implies$$
$$\left(\frac{t^2}{1-t}\right)'=-\frac{t^2-2t}{(1-t)^2}=2t+3t^2+\ldots+nt^{n-1}+\ldots=\sum_{k=1}^\infty(k+1)t^k$$