# Infinite series $\sum_{n=0}^\infty \frac{(-3)^n}{n!}$

I can show that the sum $\displaystyle \sum\limits_{n=0}^\infty \frac{(-3)^n}{n!}\;$ converges. But I do not see why the limit should be $\dfrac{1}{e^3}$.

How do I calculate the limit?

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

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Ok, that was too easy to see. I spend an hour on this, without see the result. Thanks a lot! –  leo Dec 23 '12 at 14:00
@leo It happens to best of us. –  Nameless Dec 23 '12 at 14:07
Yes, it happens to the best. But the best don't give up after just one hour. -:) –  tj_ Dec 23 '12 at 18:13
• $\quad$You'll want to remember: $\quad \displaystyle \sum_{n=0}^{\infty}\frac{x^n}{n!} = e^x$
• $\quad$For any $a, b,\;$ (provided $a\ne 0$):$\quad\displaystyle a^{-b} = \frac{1}{a^b}$