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I came across this question in my math workbook, but wasn't able to solve it.

Prove that the sequence $a_n = \frac{x^n}{n!}$ is bounded.

This is my definition of bounded: $\exists M>0 \hspace{-1mm}: \hspace{5mm} |a_n|<M \hspace{5mm} \forall n\in\mathbb{N} $.

UPDATE: I am aware I could prove convergence and then prove all convergent sequences are bounded, but I assumed there exists a direct method to prove boundedness using the definition (Hence this question was not meant as a duplicate).

Some Guy
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Xtiaan
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  • If you weren't able to solve it, what did you try? – Skam Jan 11 '18 at 20:15
  • I proved convergence, then I proved all convergent sequences are bounded. Because of the terminology in my book, I assume there is a direct method to prove boundedness. I have no clue how to continue from the definition given above. Have been searching for similar examples, but haven't found them. It is possible to prove boundedness direct, right? – Xtiaan Jan 11 '18 at 22:00

3 Answers3

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The hint.

We have $$\frac{a_{n+1}}{a_n}=\frac{x}{n+1}\rightarrow0$$

Michael Rozenberg
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Let $a_n$ be for $x \in \mathbb{R}$ the function $$ a_n\left(x\right)=\frac{x^n}{n!} $$ Hence for $x \ne 0$ $$ \left|\frac{a_{n+1}\left(x\right)}{a_n\left(x\right)}\right| \leq \frac{\left|x\right|^{n+1}}{\left(n+1\right)n!}\frac{n!}{\left|x\right|}=\frac{\left|x\right|}{n+1}\underset{n \rightarrow +\infty}{\rightarrow}0 $$ Hence the series $\sum_{n \geq 0}^{ }a_n(x)$ converges for $x \in \mathbb{R}$. Hence $$ a_n(x)\underset{n \rightarrow +\infty}{\rightarrow}0 $$ Hence, $a_n(x)$ for $x \in \mathbb{R}$ is bounded because it converges.

Atmos
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Just to be different

$\frac {x^n}{n!} = \prod \frac x1\frac x2 ....... \frac x{n}$.

Let $n > |x|$. Then $a_(n+1) = \frac {x^{n+1}}{(n+1)!} = (\prod \frac x1\frac x2 ....... \frac x{n})\frac x{n+1} < a_n$

So $a_n \le \max \{a_m\}_{m \le |x|}$.

fleablood
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