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.
Best Answer
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.