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

Question

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?

Answer 1

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

Answer 2

Hints:

• $\quad$You'll want to remember: $\quad \displaystyle \sum_{n=0}^{\infty}\frac{x^n}{n!} = e^x$

• $\quad$For any $a, b,\;$ (provided $a\ne 0$):$\quad\displaystyle a^{-b} = \frac{1}{a^b}$

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