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How do I get from $$\sum_{n=0}^{\infty}\frac{(1/e)^n}{n!} = e^{1/e}$$

I am given

$$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$$

I am thinking

$$\sum \frac{n!}{n^n}\cdot \frac{1}{n!}$$

But it seems wrong

1 Answers 1

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Let $x=1/e$.${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$

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    But then how do I get from $\sum{\frac{x^n}{n!}}=x$ then?2012-04-20
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    @JiewMeng: Huh? That is simply not true. Look at the equation $\sum \frac{x^n}{n!} = e^x$ again.2012-04-20
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    @Jiew - loosely speaking, the formula Martin pointed to says that - $\sum \frac{something^n}{n!} = e^{something}$2012-05-22