Radius of convergence of $\displaystyle{\sum_{n=0}^{\infty}} {(n!)^3 \over (3n)!}z^{3n} $.

Question

Find the radius of convergence of $\displaystyle{\sum_{n=0}^{\infty}} {(n!)^3 \over (3n)!}z^{3n} \ ?$

I applied Cauchy-Hadamard test and the result is coming $0$ (radius of convergence). To obtain the limit I also used Cauchy's first limit theorem. For a lot of messy calculation I didn't provide my work.

Please someone check whether I'm right or wrong.

Thank you..

Since $\frac{3n!}{(n!)^3}$ is an integer, all your coefficients has magnitude $\le 1$. This means the radius of convergence is at least $1$. — achille hui, Sep 05 '17 at 10:19

score 4 · Accepted Answer · answered Sep 05 '17 at 10:26

By the ratio test the sum converges if $\lim \limits_{n \to \infty} \frac{a_{n+1}}{a_n} = r$ and $|r| <1$ , special cases are when $|r|=1$ and have to be checked separately.

$a_n = \frac{(n !)^3}{(3n)!} z^{3n}$ so $\lim \limits_{n \to \infty} \frac{a_{n+1}}{a_n} =\frac{((n+1)!)^3 z^{3 n+3}}{\frac{(3 (n+1))! \left((n!)^3 z^{3 n}\right)}{(3 n)!}} = \frac{z^3 (3 n)! ((1+n)!)^3}{(n!)^3 (3 (1+n))!} = \frac{(1+n)^2 z^3}{3 (2+9 n (1+n))}=|z^3| \lim \limits_{n \to \infty}{\frac{(1+n)^2}{3(2+9n(1+n))}} = |z^3| \frac{1}{27} $

So for $|\frac{z^3}{27}| <1 $ its obvious that $-3 < z < 3$ (if you are looking for the domain of convergence for complex $z$, from here you have to continue dealing with $z$ as complex number, the methods above works for real and complex).

So when $|\frac{z^3}{27}|=1$ we get that $z=-3,3$, we need to check them.

Just substitute instead of $z$ the numbers $-3$ and then $3$ in the sum, in both cases the sum diverges using other tests than the ratio test because its inconclusive when the ratio is exactly one.

for instance $\frac{((n+1)!)^3}{(n!)^3} = \frac{(n+1)^3 *(n!)^3}{(n!)^3} = (n+1)^3$, so nothing is lost, i just factored and simplified them. — Ahmad, Sep 05 '17 at 10:36
Yes, I had just checked the details with paper and pencil. I wonder why I don't obtain the same result with Hadamard's formula. — Bernard, Sep 05 '17 at 10:39
@Ahmad If I take $a_n={(n!)^3 \over(3n)!}$ for $n=3m , m \in \mathbb N \cup { 0}$ and $a_n=0 $ for $n \neq 3m$ then we can compare the series $\displaystyle \sum_{n=0}^{\infty} a_nz^n$ with the given series.But for $\displaystyle \sum_{n=0}^{\infty} a_nz^n$ the radius of convergence is $27$ by Cauchy-Hadamard, right? — Mini_me, Sep 05 '17 at 11:26

Bernard · Answer 2 · 2017-09-05T11:21:50.587

3

Hint: First find the radius of convergence of the series $\;\displaystyle\sum_{n=0}^{\infty} {(n!)^3 \over (3n)!}z^{n}$.

For this apply Cauchy-Hadamard's formula, combined with Stirling's formula: \begin{align} \biggl(\frac{(n!)^3}{(3n)!}\biggr)^{\!\tfrac1n}\sim_\infty&\frac{\Bigl(\sqrt{(2\pi n)^{3\mathstrut}}\Bigr)^{\!\tfrac1n}\Bigl(\dfrac n{\mathrm e}\Bigr)^{\!3}}{\Bigl(\sqrt{6\pi n\mathstrut}\Bigr)^{\!\tfrac1n}\Bigl(\dfrac{3n}{\mathrm e}\Bigr)^{\!3}}=\frac{1}{27}\sqrt{\Bigl(\frac 43\pi^2n^2\Bigr)^{_\tfrac1n}}, \end{align} and we obtain a radius of convergence equal to $27$ for the auxiliary series.

edited Sep 05 '17 at 11:21

answered Sep 05 '17 at 10:27

Bernard

175,478

You're missing an exponent $n$ for the fraction in the numerator, and $3n$ in the denominator – Bart Michels Sep 05 '17 at 10:42
@barto: I've redone my calculations. Hopefully, it's correct now. – Bernard Sep 05 '17 at 10:57
In the denominator it should be $(3n/e)^3$. You may want to add an extra step in between. – Bart Michels Sep 05 '17 at 11:00
Why! Yes of course! Thanks! I'll fix it at once. A typical case where the ratio test is simpler. – Bernard Sep 05 '17 at 11:21
What does $\sim_\infty$ mean? – gen-ℤ ready to perish Sep 05 '17 at 11:37
The functions are equivalent at infinity. This means their ratio tends to $1$ at infinity. It's one of the most useful concepts for the calculation of limits (much more than L'Hospital's rule!). – Bernard Sep 05 '17 at 11:52

score 1 · Answer 3 · answered Sep 05 '17 at 11:40

If we represent the factorials everywhere as gamma-functions and Pochhammer symbols, this is actually a generalized hypergeometric function 4F3(...;;;z/3). According to wikipedia the radius of convergence is where the argument becomes 1, which means in our case, |z|=3.

Radius of convergence of $\displaystyle{\sum_{n=0}^{\infty}} {(n!)^3 \over (3n)!}z^{3n} $.

3 Answers3