Find the value of $R$ such that $\sum_{n=0}^{\infty} \frac{x^n}{n! \cdot (\ln 3)^n} = 3^x, \forall x \in (-R,R)$ I am not quite sure how to finish this exercise:


  Find the value of $R$ such that $$\sum_{n=0}^{\infty} \frac{x^n}{n! \cdot (\ln 
 3)^n} = 3^x, \forall x \in (-R,R)$$


I am honestly lost after I verify that the power series converges:

$$\begin{align*}
a_n &amp;= \frac{x^n}{n! \cdot (\ln 3)^n}\\
a_{n+1} &amp;= \frac{x^{n+1}}{(n+1)! \cdot (\ln 3)^{n+1}}
\end{align*}
$$
The ratio between the two and the limit are shown below:

$$\begin{align*}
\frac{a_{n+1}}{a_n} &amp;= \frac{x}{(n+1) \cdot \ln 3}\\
\\
\lim_{n \to \infty} \left\vert \frac{a_{n+1}}{a_n} \right\vert &amp;= \frac{\vert x \vert}{\ln 3} \cdot \lim_{n \to \infty} \frac{1}{n+1} = 0
\end{align*}
$$

So as I mentioned, I am not sure what to do next. Any guidance is highly appreciated.

Thank you.
Hint: It's possible the problem is misstated as @MathLover suggested in a comment. If it's not misstated, recall that $\sum_{n=0}^{\infty}\frac{u^n}{n!}= e^u$ and let $u=x/\ln 3.$
The $\ln 3$ does not matter.
The series converges for
all $x$.

As you have shown,
$|a_{n+1}/a_n| \to 0$.
The $\ln 3$ does 
not affect this,
so it still holds
with any constant.
This means that the series converges.