What's the limit of this expression: $\lim\limits_{M \to \infty}1/(\sum_{i=0}^{\infty}\frac{M!}{\left(M+i\right)!}x^{i})$ I have a question about a limit: Assume $x$ is a positive real constant $(x>0)$, then what's the limit of the following expression?
$$
\lim_{M\rightarrow\infty}\frac{1}{\sum_{i=0}^{\infty}\frac{M!}{\left(M+i\right)!}x^{i}}
$$
Is this dependent on the value of $x$? Thank you very much....
 A: As $1/x$ is continuous, you need to calculate
$$\lim_{M\to \infty} \sum_{i=0}^{\infty}\frac{M!}{\left(M+i\right)!}x^{i}.$$
We have
$$ \frac{1}{M^i} \geq \frac{M!}{(M+i)!} $$ 
independent of $x$. Thus,
$$\frac{M}{M-x}=\sum_{i=0}^\infty \frac{1}{M^i} x^i \geq\sum_{i=0}^{\infty}\frac{M!}{\left(M+i\right)!}x^{i} .$$
With $M\to\infty$, we find that the limit of the sum is smaller or equal to 1 for all $x$.
To find a lower bound, we just take the term corresponding to $i=0$ (all terms are positive), and we have
$$\sum_{i=0}^{\infty}\frac{M!}{\left(M+i\right)!}x^{i} \geq 1.$$
Concluding, we have that $$\lim_{M\to \infty} \sum_{i=0}^{\infty}\frac{M!}{\left(M+i\right)!}x^{i}=1$$
so your limit is also 1 independent of $x$.
A: The sum can be evaluated in closed form:
$$
   \sum_{n=0}^\infty \frac{M!}{(M+n)!} x^n = \sum_{n=0}^\infty \frac{M! n!}{(M+n)!} \frac{x^n}{n!} = M \sum_{n=0}^\infty \operatorname{B}\left(M,n+1\right) \frac{x^n}{n!}
$$ 
Using Euler's integral:
$$
   \operatorname{B}\left(M,n+1\right) = \int_0^1 (1-u)^{M-1} u^n \mathrm{d} u 
$$
and interchanging the integration and summation (warranted because of absolute convergence):
$$
  \sum_{n=0}^\infty \frac{M!}{(M+n)!} x^n = M \int_0^1 (1-u)^{M-1} \mathrm{e}^{u x} \mathrm{d} u \stackrel{\text{by parts}}{=} 1 + x \int_0^1 (1-u)^M \mathrm{e}^{u x} \mathrm{d}u 
$$
Since $\mathrm{e}^{u x} \leqslant \mathrm{e}^x$ for all $x>0$ and all $0<u<1$, the limit is easy:
$$
    0 < x \int_0^1 (1-u)^M \mathrm{e}^{u x} \mathrm{d} u < x \mathrm{e}^{x} \int_0^1 (1-u)^M \mathrm{d} u = \frac{x \mathrm{e}^{x}}{M+1}
$$
The upper bound approaches zero, hence
$$
   \lim_{M \to \infty} \sum_{n=0}^\infty \frac{M!}{(M+n)!} x^n = 1
$$
