# Radius of convergence, ratio test fails

We have $$\sum_{n=1}^{\infty} n!x^n$$

In Wolfram-ALpha it says, the series does not converge. I tried the ratio test $$lim | \frac{a_n}{a_{n+1}}|$$ and I got 0. But when I put x =0, the n! remains and that can't converge. But following my task, there should be a radius of convergence. What am I doing wrong?

• The radius of convergence is $0$. Commented Jan 24, 2017 at 20:08

The radius of convergence is $0$. Hence the power series does converge for $|x| < 0$ which is the empty set. We may check convergence on the boundary, hence for $x = 0$ but then the series trivially converges since it is constant $0$ by considering $$\sum_{n = 1}^\infty n!0^n = \sum_{n = 1}^\infty n!0 = \sum_{n = 1}^\infty 0 = 0$$

The ratio test is in fact $\lim_{n\rightarrow \infty}|\frac{a_{n+1}}{a_n}|.$

This is really $(n+1)x$, so if we want $|(n+1)x|<1$ as n goes to infinity then we must have $|x|<0$, which never happens. ($x=0$ does work but it's a boundary case and that gives a sum 0 anyway)

• I am a bit confused. There are two ratio tests, one for sequences, then your ratio test would be correct and then one for series which I have written in my question. We had this in our lecture. So therefore I am not really sure now, what to use. Commented Jan 24, 2017 at 20:12
• They should both be using the same one. It doesn't really matter; if the terms keep increasing whether it's a sequence or a series it's still going to diverge, no? Commented Jan 24, 2017 at 20:14
• @Blnpwr You may want to consider the sequence $a_n :=n! x^n$, then it is more natural to use the quotient criterion for series. Btw the quotient criterion for power series is just an adaption of the usual quotient criterion. Commented Jan 24, 2017 at 20:17
• @TheGeekGreek Yes, I considered $$an:=n!x^n$$ as a sequence. Well, not $$n!x^n$$ but $$n!$$ and then I used the ratio test for radius of convergence, but I should be using the ratio criterion for sequences. As far as I understood, when using the ratio criterion there should not be any x parameters. So our sequence here is not $$n!x^n$$ but $$n!$$, right? Commented Jan 24, 2017 at 20:25
• @Blnpwr No. There are two ways: $$\sum_{n = 1}^\infty a_n x^n \qquad \text{and} \qquad \sum_{n = 1}^\infty b_n$$ where $b_n := a_n x^n$. Now we get the convergence radius $R$ either by $$R = \lim_{n \to \infty} \left| \frac{a_{n}}{a_{n+1}}\right|$$ or by requiring $$\lim_{n \to \infty}\left|\frac{b_{n+1}}{b_n}\right| < 1$$ and solving for $|x|$ which yields the same result . Commented Jan 24, 2017 at 20:43

$n!$ will ultimately grow faster than any polynomial.

When $n>\frac 1x$ then each consecutive member in the series is larger than the one before it, and you need them to be getting smaller for the series to converge.