# Find the radius of convergence of the series $\sum a_n z^n$ when: $a_n = \frac{(n!)^3}{(3n)!}$.

Find the radius of convergence of the series $\sum a_n z^n$ when: $a_n = \frac{(n!)^3}{(3n)!}$.

I think I need to use Stirling's formula here, i.e. $n! \sim cn^{n+1/2}e^{-n}.$

But we need to find the limsup of $|a_n|^{1/n}$ here. $a_n \sim b_n$ here means that $\lim \frac{a_n}{b_n} =1$.

My question is: When we have $\lim \frac{a_n}{b_n}=1$, then how do we have $\lim a_n^{1/n} = \lim b_n^{1/n}$?

• Why not use the ratio test instead of the root test? – carmichael561 Mar 6 '18 at 22:26
• @carmichael561 I guess my real question here is how does the asymptotics pass over when we take the power of $1/n$. – nomadicmathematician Mar 6 '18 at 22:29

Compute the limit of $\frac{a_n}{a_{n+1}}$ instead.

It is easier.

$\frac{a_n}{a_{n+1}}=\frac{(3n+3)(3n+2)(3n+1)}{(n+1)^3}\to 3^3$

Since this exists then so does the limit in your formula for the radius of convergence and to the same value.

To answer your new question. If $x_n\to 1$ then $x_n^{1/n}\to 1$. Therefore, if $\lim\frac{a_n}{b_n}=1$ then $\lim \frac{a_n^{1/n}}{b_n^{1/n}}=1$. If you know in addition that $\lim b_n^{1/n}$ exists, then $\lim a_n^{1/n}=\lim b_n^{1/n}$.

Advice: When in doubt on how to replace a term by its asymptotic, do it step by step multiplying and dividing by the asymptotic.

$$a_n^{1/n}=\left(\frac{(n!/\text{Stirling}_n)^3\cdot\text{Stirling}_n^3}{(3n)!/\text{Stirling}_{3n}\cdot \text{Stirling}_{3n}}\right)^{1/n}=\left(\frac{(n!/\text{Stirling}_n)^3}{(3n)!/\text{Stirling}_{3n}}\right)^{1/n}\cdot\left(\frac{\text{Stirling}_n^3}{\text{Stirling}_{3n}}\right)^{1/n}$$ The first factor tends to $1$ and the second is your simplified expression to compute.