<s>[INST] Finding a recursive relation from a differential equation. We consider the following differential equation: $$(1-x)\psi'(x) - 1 = 0$$
We then define:  $\psi(x) = \sum_{n = 0}^{\infty}a_nx^n$.(power series)
Plugging it into the formula we get to this step: $$\sum_{n=0}^{\infty}a_nn(x^{n-1}-x^{n}) - 1= 0$$
How can we then identify the recursive relation between the summands thus determine $a_{n+1}$ ?
 [/INST] Assuming $\psi(x) = a_0 + a_1 x+a_2 x^2+\ldots$, then it follows
\begin{align}
(1-x)\psi'(x) =&\ (1-x)(a_1 + 2a_2x+\ldots + na_n x^{n-1}+\ldots)\\
=&\ a_1+ (2a_2-a_1)x + (3a_3-2a_2)x^2+\ldots + ((n+1)a_{n+1}-na_n)x^n+\ldots
\end{align}
So, we have
\begin{align}
(1-x)\psi'(x)-1 = (a_1-1) + (2a_2-a_1)x + \ldots + ((n+1)a_{n+1}-na_n)x^n+\ldots = 0
\end{align}
We know that in order for a power series to be identicially zero, then it must be that all of its coefficients are zeros. Hence we have
\begin{align}
a_1-1 = 0, 2a_2-a_1=0, \ldots, (n+1)a_{n+1}-n a_n = 0.
\end{align}
So the recursion formula is
\begin{align}
a_{n+1} = \frac{n}{n+1}a_n \ \ \ \text{ with } \ \ \ a_1=1. 
\end{align}
Solving the recursion, we see that
\begin{align}
a_n = \frac{n-1}{n}a_{n-1} = \frac{n-1}{n}\frac{n-2}{n-1}a_{n-2}= \frac{n-1}{n}\frac{n-2}{n-1}\cdots\frac{1}{2}a_1 = \frac{1}{n}a_1 = \frac{1}{n}. 
\end{align}
Thus, the solution is given by
\begin{align}
\psi(x) = a_0+\sum^\infty_{n=1} \frac{x^n}{n} = a_0 - \log(1-x)
\end{align}
for some constant $a_0$.
</s>