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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.

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    $\begingroup$ @bru1987 Are you sure that it is $\frac{x}{\ln(3)}$ instead of $x \ln(3)$? $\endgroup$
    – Math Lover
    Feb 20, 2018 at 17:27
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    $\begingroup$ You should not try to find this.Mclaren series coefficients are unique(if they exist) if you find them for right hand side they wont be equal to left.So it will not exist. $\endgroup$
    – user528935
    Feb 20, 2018 at 18:40
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    $\begingroup$ I'm not even going to do this in an aswer. By the definition of the exponential function: $$3^x=e^{ x \ln 3} = \sum_{n=0}^\infty \frac{x^n (\ln 3)^n}{n!}$$ You see the problem here? $\endgroup$
    – Yuriy S
    Feb 20, 2018 at 20:17
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    $\begingroup$ $$\frac{x}{\ln 3} \neq x \ln 3$$ unless $x=0$. This has no solution. Likely a typo as many people said $\endgroup$
    – Yuriy S
    Feb 21, 2018 at 0:53
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    $\begingroup$ If the assignment is quoted correctly, then the answer is no solution, because we can't set $R=0$ as $(0,0)$ is the empty set $\endgroup$
    – Yuriy S
    Feb 21, 2018 at 1:01

2 Answers 2

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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 $u=x/\ln 3.$

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  • $\begingroup$ I see @zhw. This exercise is not intuitive at all to me. Let me try to find out if this is a typing mistake from the teachers notes. Thank you. $\endgroup$
    – bru1987
    Feb 20, 2018 at 18:52
  • $\begingroup$ Hi @zhw. yes the description of the exercise is correct. $\endgroup$
    – bru1987
    Feb 21, 2018 at 0:41
  • $\begingroup$ @bru1987 OK, this makes it easy. There is no such $R$ as in the comments of Yuriy S $\endgroup$
    – zhw.
    Feb 21, 2018 at 1:06
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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.

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    $\begingroup$ But the question is not whether it converges or not $\endgroup$
    – Yuriy S
    Feb 20, 2018 at 20:35
  • $\begingroup$ So $R = \infty$. $\endgroup$
    – marty cohen
    Feb 20, 2018 at 22:18
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    $\begingroup$ @ martycohen, have you read the question? It is not asking for the radius of convergence. It is asking for the interval where the equation holds. Which is nowhere. Because $$\frac{x}{\log 3} \neq x \ln 3$$ unless $x=0$, but we can't write $R=0$, because then $(0,0)$ would be the empty set $\endgroup$
    – Yuriy S
    Feb 20, 2018 at 22:20

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