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I'm trying to find the radius of convergence for a series, and could use a hint.

The series is:

$$\sum_{n=1}^\infty \frac{z^n}{n!}$$

I've gotten as far as deciding that the radius will equal the reciprocal of $\limsup_{n\to \infty} \sqrt[n]\frac{1}{n!}$, if it exists. I can also see that the limit does, in fact, exist, since, $\forall n$, $\sqrt[n]\frac{1}{n!} \le \sqrt[n]\frac{1}{n}$, and the sequence $\{\sqrt[n]\frac{1}{n}\}$ converges to 1. I don't know where I should start in trying to find the exact limit, however.

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    $\begingroup$ For any $r>0$, for every $n$ big enough we have $n!>r^n$, hence the radius of convergence is $+\infty$. $\endgroup$
    – Jack D'Aurizio
    Commented Apr 22, 2016 at 0:06
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    $\begingroup$ Use the Ratio Test, more pleasant in this case than Root Test. As to finding the sum, if that is what you are looking for, it is very close to a familiar series. $\endgroup$
    – André Nicolas
    Commented Apr 22, 2016 at 0:09
  • $\begingroup$ @JackD'Aurizio That's interesting; how would you prove that, $\forall r\gt0$, $n! \gt r^n$ eventually? $\endgroup$
    – Measure Theory Penguin
    Commented Apr 22, 2016 at 0:37
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    $\begingroup$ @JohnFogg: an interesting chance is given by the identity $$\sum_{k=0}^{n}\binom{n}{k}(-1)^{k}(n-k)^n = n!$$ that comes from the theory of forward differences. It shows that $n!$ behaves like $\left(\frac{n}{e}\right)^n$, essentially, so $n>2re$ is more than enough to grant $n!>r^n$. That can be shown also through the AM-GM inequality. $\endgroup$
    – Jack D'Aurizio
    Commented Apr 22, 2016 at 0:40
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    $\begingroup$ Another chance is given by noticing that $\frac{(n+1)!}{n!}=n+1$ while $\frac{r^{n+1}}{r^{n}}=r$, so $n!$ grows faster than $r^n$ for any $r\geq 2$. $\endgroup$
    – Jack D'Aurizio
    Commented Apr 22, 2016 at 0:44

1 Answer 1

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The radius of convergence is infinite; $z$ can be as big as it wants, and the series will still converge.

As proof: if we take $s_n$ to denote an individual term in the series, the Ratio Test tells us that the series converges as long as

$$\lim_{n\to\infty}|\frac{s_{n+1}}{s_n}|<1$$

If we further note that

$$|\frac{s_{n+1}}{s_n}|=|\frac{n!}{(n+1)!}|*|\frac{z^{n+1}}{z^n}|=\frac{1}{n+1}*z$$

then the convergence of $\frac{1}{n+1}$ to $0$ and the product property of limits imply that $|\frac{s_{n+1}}{s_n}|$ converges to $0,$ and thus that the series converges, regardless of the value of $z$.

Thanks to Andre Nicolas, who suggested this solution in the comments. Thanks also to Jack D'Aurizio and user1952009, who suggested other solutions that, though less direct, are more interesting.

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