Convergence at the boundary of $\sum_{n= 1}^\infty\frac{(n!)^3}{(3n)!}z^n $

Question

We are given the complex series of functions $\sum_{n= 1}^\infty\frac{(n!)^3}{(3n)!}z^n $. It is a power series, and we can easily calculate that its disk of convergence is $D(0,27)$. How can we know if the series converges at some point of the boundary? Also, can we say to which function it converges? (I mean, can we express the sum function in an elementary way?)

Answer 1

Take some point on the boundary and then try to figure out whether the series you get by plugging in that point for $z$ converges. In the complex case you will mostly work with estimations then as you have infinitely many boundary points. Lets do it for your example:

Let $z \in \mathbb{C}$ with $\vert z \vert = 27$, i.e. $z$ lies on the boundary. Then we get

\begin{align*}\left\lvert \frac{(n!)^3}{(3n)!}z^n \right\rvert &= 27^n \cdot \frac{n}{3n} \cdot \frac{n}{3n-1} \cdot \frac{n}{3n-2} \cdot \frac{n-1}{3(n-1)} \cdot \frac{n-1}{3(n-1)-1} \cdots \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{1}{1}\\ &= 27^n \cdot \frac{1}{3} \cdot \frac{n}{3n-1} \cdot \frac{n}{3n-2} \cdot \frac{1}{3} \cdot \frac{n-1}{3(n-1)-1} \cdots \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{1}{1}\\ &= 9^n \cdot \frac{n}{3n-1} \cdot \frac{n}{3n-2} \cdot \frac{n-1}{3(n-1)-1} \cdots \frac{1}{2} \cdot \frac{1}{1}\\ &= \frac{3n}{3n-1} \cdot \frac{3n}{3n-2} \cdot \frac{3(n-1)}{3(n-1)-1} \cdots \frac{3}{2} \cdot \frac{3}{1},\end{align*} which yields $$\left\lvert \frac{(n!)^3}{(3n)!}z^n \right\rvert \geq 1,$$ as all the factors of our computation above are bigger than $1$. That means that the series cannot converge for any point on the boundary (Recall that if a series $\sum_{n = 1}^{\infty} a_n$ converges, then $a_n \rightarrow 0$ for $n \rightarrow \infty$).

Note that my argument can be generalized to the case of $\sum_{n = 1}^{\infty} \frac{(n!)^k}{(kn)!}$ for all natural numbers $k \geq 2$.

Most of the times you should not expect to be able to actually compute the value of some series. There are certainly series where this is possible (geometric ones, some generalized harmonic ones etc.), but that is generally non-trivial. As far as I know, one does not even know the value of the series $\sum_{n = 1}^{\infty}\frac{1}{n^3}$ for example.

Answer 2

Using Stirling, for $|z| = 27$: $$ \left|\frac{(n!)^3}{(3n)!}z^n\right| = \frac{(n!)^3}{(3n)!}27^n\approx \frac{2\pi n}{\sqrt 3}\not\to 0, $$ so...