# 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....

• Hint: $\frac{M!}{(M+i)!} = \frac{1}{(M+1)(M+2)\dots(M+i)} \leq \frac{1}{\left(M+1\right)^i}$ – sdcvvc Aug 23 '12 at 15:02
• Thank you~~ This is very helpful~~ ^_^ – Chang Aug 23 '12 at 16:11

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 $x$.

• Thank you very very much~~~I thought about this method before, but I got stuck...You made this very clear...Thank you... – Chang Aug 23 '12 at 16:10
• I don't understand "we find that the sum is smaller than 1 for all $x$". Certainly since $x$ is positive (and I suppose $M$ as well) $\frac M{M-x}=1+\frac x{M-x}>1$? Besides, the sum has $1$ as its first term, the others are positive. I think you can conclude that the limit of the sum is not strictly bigger than $1$, therefore $1$, at this point. – Marc van Leeuwen Sep 1 '12 at 8:18
• @MarcvanLeeuwen: smaller should mean smaller or equal. I fixed this. Thank you... – Fabian Sep 1 '12 at 12:02
• @Fabian: "limit of the sum" instead of "sum" would be even better. Since in fact the sum is always bigger than $1$, just it tends to $1$ as $M\to\infty$. Excuse me for being strict; I did notice the sequel shows you understood the situation, but statements should be correct as written. – Marc van Leeuwen Sep 1 '12 at 12:19

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 $$\lim_{M \to \infty} \sum_{n=0}^\infty \frac{M!}{(M+n)!} x^n = 1$$

• Wow, it is so cool...only one unimportant thing: I think $$M \int_0^1 (1-u)^{M-1} \mathrm{e}^{u x} \mathrm{d} u \stackrel{\text{by parts}}{=} 1 + x \int_0^1 (1-u)^M \mathrm{e}^{u x} \mathrm{d}u$$ should be plus but not the minus. Thanks soooooo much~~~ – Chang Aug 23 '12 at 16:06
• @Chang I am glad you find it useful. Thanks for pointing out the typo. I've fixed that now. – Sasha Aug 23 '12 at 16:09