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I saw at the solution of an exercise that $$\frac{|x|^{n+1}}{(n+1)!} \to 0, \text{ when n } \to +\infty$$

But,how can I show that it is actually like that?

vonbrand
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evinda
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    Let $c_k = \frac{\lvert x\rvert^k}{k!}$. Pick an $n \geqslant 2\lvert x\rvert$, and look at $c_n, c_{n+1},\dotsc$. – Daniel Fischer May 22 '14 at 15:23
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    If you write numerator and denominator as a product, how does it look like when $n> \vert x \vert$? – Roland May 22 '14 at 15:23

2 Answers2

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Take $x$ positive, that does no harm. Now consider $n \to \infty$ in: $$ \frac{x^{n + 2} / (n + 2)!}{x^{n + 1} / (n + 1)!} = \frac{x}{n + 2} \to 0 $$ As the ratio tends to zero, so does each term.

vonbrand
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Let $k\in \mathbb{N}$ such that $k + 1 > \vert x \vert$. For $n>k$, we can write the term as

$$\frac{\vert x \vert \cdot \, \dots \, \cdot \vert x \vert\cdot \vert x \vert}{1\cdot \dots \, \cdot n \cdot (n+1)}= \frac{\vert x \vert^k }{1\cdot \dots \, \cdot k}\cdot \frac{\vert x \vert^{n-k}}{(k+1)\cdot \dots \, \cdot (n+1)}$$

The first ratio in this product is some number which does not depend on $n$. The second ratio has $(n-k)$ factors in the numerator and the denominator, thus it can be written as a product of $(n-k)$ ratios:

$$\frac{\vert x \vert^{n-k}}{(k+1)\cdot \dots \, \cdot (n+1)}= \frac{\vert x \vert}{k+1}\cdot\frac{\vert x \vert}{k+2}\cdot \dots \, \cdot \frac{\vert x \vert}{n+1},$$

where each of the ratios is less than $1$ (since $k+1 > \vert x \vert$). As $n$ increases, these ratios tend to zero.

Roland
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