Let $f_0(x)=\frac{1}{1-x}$ and define $f_{n+1}(x)=xf'_n(x)$ prove that
$f_{n+1}(x)>0$ for $0<x<1$ and $n\in \mathbb{N}$
my attempt was as following. Since the expression invloves the natural numbers, induction will be a good method of proof. After evaluating few terms, I found that
$f_n(x)=\frac{g(x)}{{(1-x)}^{2n}}$
So the proof is equivelent to prove $g(x)>0$
Any hint will be appreciated
Let $\epsilon > 0$ and consider the compact set $[\epsilon, 1-\epsilon]$. Then on this set we have that the power series $$\sum_{k=0}^\infty x^k = \frac{1}{1-x}$$ and that the convergence is occurring inside the radius of convergence so we know that $$ f_1(x) = x\frac{d}{dx}\sum_{k=0}^\infty x^k = x\sum_{k=0}^\infty \frac{d(x^k)}{dx} = \sum_{k=1}^\infty kx^k$$ By induction you can show that for any $n$ that $$\sum_{k=1}^\infty k^n x^k$$ has the same radius of convergence as the original power series by using the root test. Hence we can continue to switch the order of the series and derivative. By doing this we can calculate by induction that $$f_n(x) = \sum_{k=1}^\infty k^n x^k$$ for any $x\in[\epsilon, 1-\epsilon]$. Then since $x > 0$, we can see that $f_n(x)$ is always equal to a convergent series of positive numbers. Hence $f_n(x) > 0$ for all $x\in[\epsilon, 1-\epsilon]$. You can show this holds for all $\epsilon >0$, hence the same conclusion holds on $(0,1)$ as desired.