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

Question

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:

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

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

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

Thank you.

Answer 1

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 $u=x/\ln 3.$

Answer 2

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.