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: $$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$$
Nameless
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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
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3@leo It happens to best of us. – Nameless Dec 23 '12 at 14:07
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2Yes, it happens to the best. But the best don't give up after just one hour. -:) – tj_ Dec 23 '12 at 18:13
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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 $a\ne 0$):$\quad\displaystyle a^{-b} = \frac{1}{a^b}$
amWhy
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Thank you! A helpful addition to the answer of @Nameless. But I did know this fact already. – leo Dec 23 '12 at 14:12
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2@Sigur - take heart! rep is just a number; my posts now are not much different (in terms of quality) than my posts when I was below 2k. – amWhy Dec 23 '12 at 14:56