<s>[INST] Radius of convergence opposite result I am trying to find the radius of convergence of the series $\sum_{n=1}^\infty{\frac{(n!)^3 \cdot x^n}{(3n)!}}$. I know that it should be $|x|<27$, but i get the exactly the opposite result ($27<|x|$). I am using the ratio test to determine it (e.g. $\rho = \lim_{n\to\infty} \left|\frac{a_n}{a_{n+1}}\right|$).
My calculations are as follows:
$$
\lim_{n\to\infty} \left|\frac{\frac{(n!)^3 \cdot x^n}{(3n)!}}{\frac{((n+1)!)^3 \cdot x^{n+1}}{(3(n+1))!}} \right|
= \left| \lim_{n\to \infty} \frac{(n!)^3 \cdot x^n}{(3n)! \cdot \frac{((n+1)!)^3 \cdot x^{n+1}}{(3(n+1))!}} \right|
=\lim_{n \to \infty} \left| \frac{\frac{(n!)^3 \cdot x^n}{\frac{((n+1)!)^3 \cdot x^{n+1}}{(3(n+1))!}}}{(3n)!} \right|
= \lim_{n\to\infty} \left| \frac{\frac{(n!)^3 \cdot x^n \cdot (3(n+1))!}{((n+1)!)^3 \cdot x^{n+1}}}{(3n)!} \right|
= \lim_{n\to \infty} \left| \frac{(n!)^3 \cdot x^n \cdot (3(n+1))!}{((n+1)!)^3 \cdot x^{n+1} \cdot (3n)!} \right|
= \lim_{n\to\infty} \left| \frac{(n!)^3 \cdot (3(n+1))!}{((n+1)!)^3 \cdot x \cdot (3n)!} \right|
= \lim_{n\to\infty} \left| \frac{(n!)^3 \cdot (3n+3)!}{((n+1)!)^3 \cdot x \cdot (3n)!} \right|\\
= \lim_{n\to\infty} \left| \frac{(n!)^3 \cdot (3n+3)\cdot(3n+2)\cdot(3n+1)}{((n+1)!)^3 \cdot x} \right|
= \lim_{n\to\infty} \left| \frac{(3n+3)\cdot(3n+2)\cdot(3n+1)}{(n+1)^3 \cdot x} \right|\\
= \lim_{n\to\infty} \left| \frac{27 n^3 + 54 n^2 + 33 n + 6}{n^3x + 3 n^2x + 3 nx + 1x} \right|\\
\overset{\cdot \frac{\frac{1}{n^3}}{\frac{1}{n^3}}}{=} \lim_{n\to\infty} \left| \frac{27 + \frac{54}{n} + \frac{33}{n^2} + \frac{6}{n^3}}{x + \frac{3x}{n} + \frac{3x}{n^2} + \frac{x}{n^3}} \right|\\
\Rightarrow \frac{27}{|x|} < 1 \Rightarrow 27 < |x|
$$
Where did I go wrong? Did I somehow mess up the direction of the inequality?
 [/INST] There shouldn't be any $x$'s.  Then you will get $r=\lim\dfrac{ \mid a_n\mid}{\mid a_{n+1}\mid}=27$.
That's you use the $a_n$ in $$\sum a_nx^n$$.
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