Assume that $a_0=3$, $a_1=5$, and, for arbitrary $n>1$ , $na_n=\frac{2}{3}a_{n-1}-(n-1)a_{n-1}$. Prove that, when $|x|<1$, the series $\sum\limits_{n=0}^\infty a_nx^n$ converges, and compute its sum.
I tried to let $\displaystyle a_n-a_1=\sum_{k=2}^{n}\frac{5-6k}{3k}a_{n-1}$ , and $\displaystyle a_n=(\frac{5-3n}{3n})a_{n-1}$
$$ a_{n-1}(\frac{5-3n}{3n}-\sum_{k=2}^{n}\frac{5-6k}{3k})=5 $$
I want to know how to continue it.
Edit: (after reading ideas by @JV.Stalker)
I made the following supplement
$$ S(x)=\sum_{n=0}^{\infty}a_nx^n\\ S'(x)=\sum_{n=1}^{\infty}na_nx^{n-1}\\ \sum_{n=2}^{\infty}na_nx^n=\sum_{n=2}^{\infty}\frac{2}{3}a_{n-1}x^n-\sum_{n=2}^{\infty}(n-1)a_{n-1}x^n\\ [xS'(x)-5x]=\frac{2}{3}x·\sum_{n=2}^{\infty}(n-1)a_{n-1}x^{n-1}-x\sum_{n=2}^{\infty}(n-1)a_{n-1}x^{n-1}\\ x[S'(x)-5]=\frac{2}{3}x(S(x)-3)-x(xS'(x))\\ S'(x)-5=S(x)-3-xS'(x)\\ (x+1)S'(x)=\frac{2}{3}S(x)+3\\ S'(x)-\frac{2}{3}\frac{1}{x+1}S(x)=\frac{3}{x+1}\\ S(x)=c(x+1)^{\frac{2}{3}}-\frac{9}{2}\\ S(0)=a_0=3\\ c=\frac{15}{2}\\ S(x)=\frac{15}{2}(x+1)^{\frac{2}{3}}-\frac{9}{2} $$
$na_n=\frac{5}{3}a_{n-1}-na_{n-1}$
Multiply by $x^n$ both sides and sum from $n=1$ to $\infty$
$\sum\limits_{n=1}^\infty na_n x^n=\frac{5}{3}\sum\limits_{n=1}^\infty a_{n-1}x^n-\sum\limits_{n=1}^\infty na_{n-1}x^n$
Reindex of the RHS:
$\sum\limits_{n=1}^\infty na_nx^n=\frac{5}{3}x\sum\limits_{n=0}^\infty a_{n}x^n-x\sum\limits_{n=0}^\infty (n+1)a_{n}x^n$
After sorting the eqution:
$\sum\limits_{n=1}^\infty na_nx^n+x\sum\limits_{n=0}^\infty na_{n}x^n=\frac{2}{3}x\sum\limits_{n=0}^\infty a_{n}x^n$
$(1+x)\sum\limits_{n=0}^\infty na_nx^n=\frac{2}{3}x\sum\limits_{n=0}^\infty a_{n}x^n$
Use that $nx^{n-1}=\frac{dx^n}{dx}$ and let $f(x)=\sum\limits_{n=0}^\infty a_nx^n$
We have the following differential eqution:
$(x+1)\frac{df(x)}{dx}=\frac{2}{3}f(x)$
Finally $f(x)=(x+1)^\frac{2}{3}+c$
$\sum\limits_{n=1}^\infty a_n x^n$ is convergent.