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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?

2 Answers 2

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

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    Yes, it happens to the best. But the best don't give up after just one hour. -:)2012-12-23
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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 $a\ne 0$):$\quad\displaystyle a^{-b} = \frac{1}{a^b}$


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    @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.2012-12-23