MHB -z.54 find the radius of convergence

[karush]

$\tiny{10.7.37}$
$\displaystyle\sum_{n=1}^{\infty}
\frac{6\cdot 12 \cdot 18 \cdots 6n}{n!} x^n$
find the radius of convergence
I put 6 but that wasn't the answer

 

[Euge]

Please explain how you originally obtained $6$ as the answer.

 

[karush]

the ans was 1/6
Looked at an example very close to this
and noticed the first term revealed the answer but couldn't follow all the steps they had to get it.

 

[Euge]

Well, the coefficient $a_n$ of $x^n$ in the power series reduces to $6^n$, for $6\cdot 12\cdot 18\cdots 6n = (6\cdot 1)(6\cdot 2)\cdots (6\cdot n) = 6^nn!$. So, $\sqrt[n]{a_n} = 6$, and the radius $R$ of convergence of the power series is given by $1/\lim\limits_{n\to \infty} \sqrt[n]{a_n} = 1/6$.