Radius of convergence of fast converging power series Suppose $a_i\ge 0$ and $a_{n+1}+a_{n+2}+\cdots < 1/n!$. What can I say about the radius of convergence of
$$\operatorname{f}(x) = \sum_{n \ge 0} a_n x^n$$
The above condition gives that $\operatorname{f}(1)$ exists so that gives lower limits on the radius of convergence. 
But what is the best that can be said?
 A: Since $a_{n+1}<1/n!$, then $\sum_na_n|x|^n\le a_0+|x|\sum_n |x|^n/n!$, which converges. Hence, the series converges absolutely (and therefore converges) for all $x$. The radius of convergence is $+\infty$.
A: For any $x$, when $n\gt 2|x|$, we have
$$
\frac{\frac{|x|^{n+2}}{(n+1)!}}{\frac{|x|^{n+1}}{n!}}=\frac{|x|}{n+1}<\frac12
$$
Therefore, by the ratio test,
$$
\sum_{n=1}^\infty\frac{x^{n+1}}{n!}
$$
converges absolutely. Since $a_{n+1}\le\frac1{n!}$, we must have that
$$
\sum_{n=0}^\infty a_{n+1}\,x^{n+1}
$$
also converges absolutely by the comparison test.
Since this is true for any $x$, the radius of convergence is $\infty$.
A: We could use Stirling's formula, but it's enough to use this lemma.
Lemma: $\displaystyle \sqrt[n]{n!}>\frac{n}{e}$
Proof: Induction on $n$.  For $n=1$, $\sqrt[n]{n!}=1>\frac{1}{e}=\frac{n}{e}$.  Otherwise we assume $\sqrt[n]{n!}>\frac{n}{e}$.  Raise both sides to the $n$ power we get $n!>(\frac{n}{e})^n$.  Multiply both sides by $(n+1)$ and we get $(n+1)!>(\frac{n}{e})^n(n+1)$.  It now suffices to prove that $(\frac{n}{e})^n(n+1) > (\frac{n+1}{e})^{n+1}$, which rearranges to $e>(\frac{n+1}{n})^n=(1+\frac{1}{n})^n$, a well-known identity from calculus.
We may rearrange the lemma to get $\frac{1}{\sqrt[n]{n!}}<\frac{e}{n}$, and pull off one piece of the factorial to get $\frac{1}{\sqrt[n]{(n-1)!}}<\frac{e\sqrt[n]{n}}{n}=en^{1/n-1}$
Now, your conditions imply that $a_{n+1}<\frac{1}{n!}$, i.e. $a_n<\frac{1}{(n-1)!}$.  We can use the root test on $f(x)$ to get $\sqrt[n]{a_n|x|^n}<\frac{|x|}{\sqrt[n]{(n-1)!}}<|x|en^{1/n-1}$.  As $n\rightarrow \infty$, $|x|en^{1/n}$ is bounded, while $n^{-1}\rightarrow 0$.  Hence your series has infinite radius of convergence.
