Radius of convergence of $\sum_{k = 0}^{\infty}\frac{ k^{2 k + 5} \ln^{10} k \ln \ln k}{\left(k!\right)^2} x^k$ I want to find the radius of convergence of 
$$\sum_{k = 0}^{\infty}\frac{ k^{2 k + 5} \ln^{10} k \ln \ln k}{\left(k!\right)^2} \,x^k$$
I know formulae 
$$R=\dfrac{1}{\displaystyle\limsup_{k\to\infty} \sqrt[k]{\left\lvert a_k\right\rvert}}.$$
For this power series 
$$R= \dfrac{1}{\limsup_{k\to\infty}{\displaystyle\sqrt[k]{\dfrac{ k^{2 k + 5} \ln^{10} k \ln \ln k}{\left(k!\right)^2}}}}.$$
But I don't know how calculate $\;\displaystyle\limsup_{k\to\infty} \sqrt[k]{\frac{ k^{2 k + 5} \ln^{10} k \ln \ln k}{\left(k!\right)^2}}$
Thank you for any help!
 A: HINT:
Stirling's Formula states
$$k! =\sqrt{2\pi k}\left(\frac ke\right)^k \left(1+O\left(\frac1k\right)\right)$$
Then, 
$$\left(\frac{k^{2k+5}\,(\log k)^{10}\,\log (\log k)}{(k!)^2}\right)^{1/k}\sim e^2\,\left(\frac{k^{4}\,(\log k)^{10}\,\log (\log k)}{2\pi}\right)^{1/k}\to e^2$$
A: We have $$\sqrt[k]{\frac{ k^{2 k + 5} \ln^{10} k \ln \ln k}{\left(k!\right)^2}} = \left(\frac {k^4 e^{2k} \log^{10} k \log \log k} {2 \pi} \left(1 + O \left(\frac {1} {k}\right)\right)\right)^{1/k} \to e^2,$$ so $$R = \frac {1} {e^2}.$$
A: You can use 
$$ R=\lim_{k\to\infty}\frac{a_k}{a_{k+1}}. $$
In fact,
\begin{eqnarray}
R&=&\lim_{k\to\infty}\frac{ k^{2 k + 5} \ln^{10} k \ln \ln k}{\left(k!\right)^2}\frac{\left((k+1)!\right)^2}{ (k+1)^{2 (k+1) + 5} \ln^{10} (k+1) \ln \ln (k+1)}\\
&=&\lim_{k\to\infty}\frac{k^{2 k + 5}}{(k+1)^{2 (k+1) + 5}}\bigg[\frac{\ln k}{\ln(k+1)}\bigg]^{10}\frac{\ln\ln k}{\ln\ln(k+1)}\bigg[\frac{(k+1)!}{k!}\bigg]^2\\
&=&\lim_{k\to\infty}\frac{k^{2 k + 6}}{(k+1)^{2 (k+1) + 4}}\\
&=&\lim_{k\to\infty}\bigg(\frac{k}{k+1}\bigg)^{2k+6}\\
&=&\frac{1}{e^2}.
\end{eqnarray}
