Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


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

share|cite|improve this question
up vote 4 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 $\lim f(n) = e$.

share|cite|improve this answer
@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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.