I was trying to find the radius of convergence for $\sum_{k=1}^{\infty} {{k^33^{-k!}}(x-3)^{k!}}$.
First I rewrote the sum to $\sum_{k=1}^{\infty} {\frac{{k^3}(x-3)^{k!}}{3^{k!}}}$ and used the ratio test: $\lim_{k \to \infty}|\frac{{(k+1)^3}(x-3)^{(k+1)!}3^{k!}}{k^3(x-3)^{k!}3^{(k+1)!}}|$.
Now we can write this as: $\lim_{k \to \infty}|\frac{{(k+1)^3}(x-3)^{(k+1)!-k!}3^{k!-(k+1)!}}{k^3}|$.
Here we can solve $\lim_{k \to \infty}|\frac{(k+1)^3}{k^3}| = |\frac{{(k(1+\frac{1}{k}))}^3}{k^3}| = |\frac{k^3(1+\frac{1}{k})^3}{k^3}| = 1$.
Thus, we now have to solve: $\lim_{k \to \infty}|{(x-3)^{(k+1)!-k!}}\cdot{3^{k!-(k+1)!}}|$.
We can further use the fact that $(k+1)!-k! = k \cdot k!$ and $k!-(k+1)! = -k \cdot k!$.
Hence, $\lim_{k \to \infty}|{(x-3)^{kk!}}\cdot{3^{-kk!}}| = |{(x-3)^{kk!}}\cdot\frac{1}{3^{kk!}}| = |\frac{(x-3)^{kk!}}{3^{kk!}}| = |(\frac{{(x-3)}}{{3}})^{kk!}|$.
Obviously, if $|x-3| > 3$ this will diverge. This happens if $x > 6$ or $x < 0$. That's why I concluded that $L = \infty > 1$ if and only if $x \neq [0,...,6]$. Therefore, $R = 0$ and Interval of convergence: $0 < x < 6$.
Is this solution completely wrong? It feels like maybe I made a mistake somewhere and Wolfram Alpha couldn't compute this. I'm very happy about every help I can get or answer!
When you calculate the radius of convergence of a power series you can just work with the coefficients; values of $x$ don't come into the argument. Let $a_n=k^{3}3^{-k!}$ if $n=k!$ and $0$ otherwise. Then the given series is $\sum a_n (x-3)^{n}$. The general formula for radius $R$ of convergence of the series is given by $\frac 1 R=\lim \sup |a_n|^{1/n}$. Hence $\frac 1 R =\lim \sup_k k^{3/k!} 3^{-1}=\frac 1 3$. Hence $R=3$. (The domain of convergence is $\{x:|x-3|<3\}$ but you have only been asked to find the radius of convergence).