# Show that $(1+ \frac{1}{n})^n$ and $(1- \frac{1}{n})^{-n}$ have the same limit

Let $x$ be positive and $$a_n = \left( 1 + \frac{x}{n} \right)^n \qquad b_n = \left( 1 - \frac{x}{n} \right)^{-n}.$$ Show that

a) The sequence $(a_n)$ and $(b_n)$ have the same limit $\xi =: \operatorname{Exp}(x)$. Use the following steps

i) $a_n < b_n$ for all $n \in \mathbb{N}$ with $n > x$.

ii) for $n > x$, $(a_n)$ is monotone increasing, and $(b_n)$ monotone decreasing.

iii) $b_n - a_n \to 0$ for $n \to \infty$.

b) For $y = -x < 0$ $$\lim_{n\to \infty}\left( 1 + \frac{y}{n} \right)^n = \frac{1}{Exp(x)}$$

Item ii) is simple, because $0 < \frac{x}{n} < 1$, obvisouly $a_n$ increasing. But for the rest I have no idea...

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Need to add the homework tag –  Arjang Feb 25 '13 at 0:17
ii) is not as simple as you think. $1+(x/n)$ is decreasing, so why is $(1+(x/n))^n$ increasing? –  Gerry Myerson Feb 25 '13 at 0:18
because it is bigger than one, and multiplying two numbers bigger than one yields a bigger number then each of the two. –  Stefan Feb 25 '13 at 1:38

For (i), we have $$\frac{a_n}{b_n} = \left(1 - \frac{x^2}{n^2}\right)^n < 1$$ provided $n > x$.