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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....

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    Thank you~~ This is very helpful~~ ^_^2012-08-23

2 Answers 2

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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 $x$.

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    @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.2012-09-01
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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, 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 $

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    @Chang I am glad you find it useful. Thanks for pointing out the typo. I've fixed that now.2012-08-23