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Evaluate

$$\lim_{n\to\infty}\left(\frac{n \cdot n!}{e\cdot (-2)^{n+1}}\cdot \left(1-e^2 \sum_{k=0}^{n} \frac{(-2)^k}{k!}\right)\right)$$

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up vote 3 down vote accepted

$$\sum_{k=0}^{n} \dfrac{(-2)^k}{k!} = e^{-2} - \left( \sum_{k=n+1}^{\infty} \dfrac{(-2)^k}{k!}\right)$$ Hence, $$1 - e^2 \left(\sum_{k=0}^{n} \dfrac{(-2)^k}{k!} \right) = e^2 \left( \sum_{k=n+1}^{\infty} \dfrac{(-2)^k}{k!}\right)$$ Hence, $$f(n) = \left(\frac{n \cdot n!}{e\cdot (-2)^{n+1}}\cdot \left(1-e^2 \sum_{k=0}^{n} \frac{(-2)^k}{k!}\right)\right) = \left(\frac{n \cdot n!}{e\cdot (-2)^{n+1}}\cdot e^2 \left( \sum_{k=n+1}^{\infty} \dfrac{(-2)^k}{k!}\right)\right)$$ $$f(n) = e n \cdot n! \left( \sum_{k=0}^{\infty} \dfrac{(-2)^k}{(k+n+1)!}\right) = \dfrac{n}{n+1} \underbrace{e \cdot (n+1)! \left( \sum_{k=0}^{\infty} \dfrac{(-2)^k}{(k+n+1)!}\right)}_{g(n)}$$ $$g(n) = e \left(1 + \dfrac{(-2)}{(n+2)} + \dfrac{(-2)^2}{(n+2)(n+3)} + \cdots \right)$$ Hence, $\lim g(n) = e$ and hence $\lim f(n) = e$.

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@Chris'ssister What trick exactly are you talking about? – user17762 Nov 21 '12 at 19:51
    
@Chris'ssister I recognized that $$\sum_{k=0}^{n} \frac{(-2)^k}{k!}$$ was the leading order term of $e^{-2}$. I don't think I would categorize it as a trick. More of recalling it from memory. – user17762 Nov 21 '12 at 19:56
    
I emphasized a different thing and not the fact that $\sum_{k=0}^{n} \frac{(-2)^k}{k!}$ is the leading order term of $e^{-2}$. Anyway. – user 1618033 Nov 21 '12 at 20:15

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