Find the radius and interval of convergence of the series $\sum\limits _{n=0}^{\infty} n!x^n$

Question

I used ratio test and ended up with $|x|\lim \limits _{n\rightarrow \infty}(n+1)$

If $x\neq 0$, the limit is infinite and thus the series diverges. Now, when $x=0$, I have the series $\sum\limits _{n=0}^{\infty} n!0^n$

I am sure that all terms will be zero except the first term with $0^0$. I know this to be one of the indeterminate forms and I am not sure of what to do with it.

score 0 · Answer 1 · answered Oct 21 '19 at 09:59

0

In standard notations the starting term of the series is $(0!)(1)=1$. The series converges only for $x=0$.

answered Oct 21 '19 at 09:59

Kavi Rama Murthy

311,013

Could you please convince me that $0^0=1$? – mamotebang Oct 21 '19 at 10:02
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When we write $ \sum\limits_{n=0}^{\infty} a_nx^{n}$ we always mean $a_0+a_1x+a_2x^{2}+...$. The series starts with the constant $a_0$. This is a matter of convention. @mamotebang – Kavi Rama Murthy Oct 21 '19 at 10:04

score 0 · Accepted Answer · answered Oct 21 '19 at 10:04

Don't worry about $0^0$. We call it an indeterminate form because functions $f,\,g$ of $x\to0$ limits $0,\,0$ have no predictable $x\to0$ behaviour for $f^g$, defined as $\exp(g\ln f)$. But in $\sum_{n\ge0}a_nx^n$, the exponentiation $x^n$ is instead shorthand for $\prod_{j=0}^{n-1}x$, so $x^0$ is an empty product, i.e. $1$ as per @KaboMurphy's answer. You will similarly see people write degree-$d$ polynomials as $\sum_{k=0}^da_kx^k$, where with this convention the constant term is $a_0x^0=a_0$ for all $x$, $0$ included. So, in the case at hand, $\sum_{n\ge0}n!0^n=0!=1$.

Find the radius and interval of convergence of the series $\sum\limits _{n=0}^{\infty} n!x^n$

2 Answers2