# 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 $0$. – Bernard Jan 24 '17 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. – Blnpwr Jan 24 '17 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? – pie314271 Jan 24 '17 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. – TheGeekGreek Jan 24 '17 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? – Blnpwr Jan 24 '17 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 . – TheGeekGreek Jan 24 '17 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.