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

Question

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

Kuku · Accepted Answer · 2012-04-20 10:16:32Z

up vote 6 down vote accepted

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

answered Apr 20 '12 at 10:16

Kuku
50518

But then how do I get from $\sum{\frac{x^n}{n!}}=x$ then? – Jiew Meng Apr 20 '12 at 10:23

1

@JiewMeng: Huh? That is simply not true. Look at the equation $\sum \frac{x^n}{n!} = e^x$ again. – Martin Wanvik Apr 20 '12 at 10:28

@Jiew - loosely speaking, the formula Martin pointed to says that - $\sum \frac{something^n}{n!} = e^{something}$ – Amihai Zivan May 22 '12 at 6:45

1 Answer