convergence radius of two specific power series

Question

I did two calculations that I think are wrong but I am not sure why.

I have to compute the convergence radius of the following power series

a) $\sum_0^{\infty} \ln(k!)x^k$,

b) $\sum_0^{\infty} k8^kx^{3k}$.

Here's my attempt:

a)

$$ \limsup_{k \rightarrow \infty} \sqrt[k]{a_k} = \limsup_{k \rightarrow \infty} \sqrt[k]{\ln(k!)} = \ln (\limsup_{k \rightarrow \infty} \sqrt[k]{k!})= \infty. $$

Therefore the convergence radius should be $0$ and the series only converges for the value $x=0$. Unfortunately, Wolfram alpha gives me another answer. Where's the mistake ?

b)

$$ \limsup_{k \rightarrow \infty} \sqrt[k]{a_k} = \limsup_{k \rightarrow \infty} \sqrt[k]{k8^k} = \limsup_{k \rightarrow \infty} \sqrt[k]{k} \sqrt[k]{8^k} = 8 \limsup_{k \rightarrow \infty} \sqrt[k]{k} = 8. $$

I would now conclude that the convergence radius is $\frac{1}{8}$, but it appears to be $\frac{1}{2}$, what did I miss ?

Thank you for your help.

Answer 1

Answer for b). What you have calculated is the radius of convergence of $\sum k8^{x}x^{k}$. The given series is convergent if $|x^{3}| < \frac 1 8$ and divergent for $|x^{3}| >\frac 1 8$. Can you see from this that the radius of convergence is $\frac 1 2$?.

Part a). It is not true that $(k!)^{1/k} \to \infty$ as $k \to \infty$. Use Stirling's approximation for this part.

Answer 2

For a) you have made the error of assume the log and the exponent commute. That is $$\ln \sqrt[k]{k!} \ne \sqrt[k]{ \ln k!}$$ and hence $$ \limsup_{k \to \infty} \sqrt[k]{ \ln k!} \ne \ln \left( \limsup_{n \to \infty} \sqrt[k]{k!} \right)$$ I have to disagree with the other answer on how to evaluate this limit (after the error). Using Stirling's approximation gives $$ \sqrt[k]{k!} \sim \sqrt[k]{ \sqrt{2 \pi k} \left( \frac{k}{e} \right)^k } = \left(2 \pi k \right)^{\frac{1}{2k}} \frac{k}{e} \to \infty$$

Of course the actual problem was the one I pointed out before. Here's how I would attack it. $$\lim_{k \to \infty} \sqrt[k]{ \ln k!}=\lim_{k \to \infty} \frac{ \ln (k+1)!}{\ln (k!)}=\lim_{k \to \infty} \frac{\sum_{j=0}^{k+1} \ln j}{\sum_{j=0}^{k} \ln j}=\lim_{k \to \infty} \frac{\ln (k+1)}{\ln k}=1$$ Note I used the Stolz–Cesàro theorem in the third equality.

I agree with the other answer for (b). You missed the $3k$ in the exponent