Find $\sum_{n=1}^{\infty}a_nx^n$ given $a_0=3, \ 3na_n+3(n-1)a_{n-1}=2a_{n-1}$ Given $\ a_0=3$, $\,3na_n+3(n-1)a_{n-1}=2a_{n-1}$, find $\ f = \sum_{n=1}^{\infty}a_nx^n$.
I have proved that when $\ \left\lvert x \right\rvert<1$, this exponential series function is convergent. 
It does not  seem to be possible to solve $a_n$ as a general function. I have no idea about the first step. 
 A: Setting
$$
h(x)=a_0+\sum_{n=1}^\infty a_nx^n,
$$
we have
\begin{eqnarray}
3h'(x)&=&\sum_{n=1}^\infty 3na_nx^{n-1}=\sum_{n=1}^\infty[2a_{n-1}-3(n-1)a_{n-1}]x^{n-1}=2\sum_{n=1}^\infty a_{n-1}x^{n-1}-3\sum_{n=1}^\infty (n-1)a_{n-1}x^{n-1}\\
&=&2\sum_{n=0}^\infty a_nx^n-3\sum_{n=0}^\infty na_nx^n=2\sum_{n=0}^\infty a_nx^n-3x\sum_{n=1}^\infty na_nx^{n-1}=2h(x)-3xh'(x)
\end{eqnarray}
i.e.
$$
3(x+1)h'(x)=2h(x),\quad h(0)=a_0=3.
$$
It follows that
$$
\frac{h'(x)}{h(x)}=\frac23\cdot\frac{1}{x+1}, \quad h(0)=3.
$$
Integrating the previous IVB, we get:
$$
\ln|h(x)|-\ln3=\frac23\ln|x+1|=\ln(x+1)^{2/3},
$$
i.e.
$$
h(x)=\pm\exp\left[\ln(3(x+1)^{2/3})\right]=\pm3(x+1)^{2/3}
$$
Thus
$$\tag{1}
\sum_{n=1}^\infty a_nx^n=-a_0+h(x)=-3\pm3(x+1)^{2/3}.
$$

Since the LHS of (1) vanishes at $x=0$, we have
$$
\sum_{n=1}^\infty a_nx^n=-3+3(x+1)^{2/3}.
$$
A: Let
$f(x) = \sum_{n=1}^{\infty}a_nx^n
$.
What is needed
is to get terms in the power series
that are the terms in the
recurrence.
Those are
$n a_n$,
$(n-1)a_{n-1}$,
and $a_{n-1}$.
The operations that
are typically used
and differentiation,
integration,
and multiplying or dividing
by a power of $x$.
To get just $a_{n-1}$,
multiply by $x$.
Then
$xf(x)
=x\sum_{n=1}^{\infty}a_nx^n
=\sum_{n=1}^{\infty}a_nx^{n+1}
=\sum_{n=2}^{\infty}a_{n-1}x^n
$.
Adding in the $a_0$ term,
$xf(x)+a_0x
=\sum_{n=1}^{\infty}a_{n-1}x^n
$.
To get terms with
$na_n$,
differentiate
and multiply by $x$.
Here's why:
$f'(x)
=\sum_{n=1}^{\infty}a_nnx^{n-1}
$,
so
$xf'(x)
=\sum_{n=1}^{\infty}a_nnx^n
$.
Finally,
to get the series
with terms $(n-1)a_{n-1}
$,
take the series
with $na_n$
and multiply by $x$.
This gives
$x^2f'(x)
=\sum_{n=1}^{\infty}a_nnx^{n+1}
=\sum_{n=2}^{\infty}a_{n-1}(n-1)x^{n+1}
=\sum_{n=1}^{\infty}a_{n-1}(n-1)x^{n+1}
$.
Since
$0
=3na_n+3(n-1)a_{n-1}-2a_{n-1}
$,
$\begin{array}\\
0
&=\sum_{n=1}^{\infty} (3na_n+3(n-1)a_{n-1}-2a_{n-1})x^n\\
&=3\sum_{n=1}^{\infty} na_nx^n+3\sum_{n=1}^{\infty}(n-1)a_{n-1}x^n-2\sum_{n=1}^{\infty}a_{n-1}x^n\\
&=3xf'(x)+3x^2f'(x)-2(xf(x)+a_0x)\\
&=3xf'(x)(1+x)-2xf(x)-2a_0x\\
\end{array}
$
This
(modulo any errors I may have comitted)
is a differential equation
for $f$
which you can now proceed to solve.
