0
$\begingroup$

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 $\lim a_n^{1/n} = \lim b_n^{1/n}$?

$\endgroup$
  • 1
    $\begingroup$ Why not use the ratio test instead of the root test? $\endgroup$ – carmichael561 Mar 6 '18 at 22:26
  • $\begingroup$ @carmichael561 I guess my real question here is how does the asymptotics pass over when we take the power of $1/n$. $\endgroup$ – nomadicmathematician Mar 6 '18 at 22:29
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.