Proof that $\lim\limits_{n \to \infty} (1-\frac{2}{n})^n$ exists I know the proof that $\lim\limits_{n \to \infty} (1+\frac{1}{n})^n$ exists, but I don't think I can use the same proof using the binomial theorem for this because I get $2^i$ in there and that increases without bound. Is there another method for proving this?
 A: To show that the sequence $a_n=\left(1-\frac2n\right)^n$ converges, we will show that it is monotonically increasing and bounded above.
To show that it is monotonically increasing we analyze the ratio $\frac{a_{n+1}}{a_n}$.  To that end, we write for $n\ge 2$
$$\begin{align}
\frac{a_{n+1}}{a_n}&=\frac{\left(1-\frac2{n+1}\right)^{n+1}}{\left(1-\frac2n\right)^n}\\\\
&=\left(1+\frac{2}{(n-2)(n+1)}\right)^{n+1}\left(1-\frac2n\right) \tag 1\\\\
&\ge \left(1+\frac{2}{n-2}\right)\left(1-\frac2n\right) \tag 2\\\\
&=1
\end{align}$$
Thus, $a_{n+1}\ge a_n$ for $n\ge 2$.  In going from $(1)$ to $(2)$, we made use of Bernoulli's Inequality.
Now, it is obvious that $a_n=\left(1-\frac2n\right)^n\le 1$.  And finaly, since $a_n$ is monotonically increasing and bounded above by $1$, the sequence converges.
A: The usual trick is to write that $\left(1+\frac{a}{n}\right)^n=\left(1+\frac{1}{m}\right)^{am}$ with $m=\frac{n}{a}$, 
so that $\left(1+\frac{a}{n}\right)^n=\left(\left(1+\frac{1}{m}\right)^{m}\right)^a$. 
When $n$ approaches $\infty$, then $m$ approaches $\pm\infty$. Using that $\displaystyle{\lim_{m\to\pm\infty}\left(1+\frac{1}{m}\right)^m}$ exists and its value is $e$ (cf. below), we conclude that $\displaystyle{ \lim_{n\to\infty}\left(1+\frac{a}{n}\right)^n }$ exists and is equal to $e^a$ for any $a\in\mathbb R$.
Proof that $\displaystyle{\lim_{n\to -\infty}\left(1+\frac{1}{n}\right)^n=e}$:


*

*We first show that $\displaystyle{\lim_{n\to\infty}\left(1-\frac{1}{n}\right)^n=e^{-1}}$. Indeed, we have $$\left(1-\frac{1}{n}\right)^n=\left(\frac{n-1}{n}\right)^n=\left(1+\frac{1}{m-1}\right)^{-m}=\left[\left(1+\frac{1}{m-1}\right)^{m-1}\left(1+\frac{1}{m-1}\right)\right]^{-1}$$ so $\displaystyle{\lim_{n\to\infty}\left(1-\frac{1}{n}\right)^n=e^{-1}}$ as claimed.

*Now, $\displaystyle{\lim_{n\to -\infty}\left(1+\frac{1}{n}\right)^n=\lim_{k\to\infty}\left(1-\frac{1}{k}\right)^{-k}=\lim_{k\to\infty}\frac{1}{\left(1-\frac{1}{k}\right)^k}=\frac{1}{e^{-1}}=e}$

A: For n > 2.  $0 < 1 - 2/n < 1 + n$ so $0< (1 - 2/n)^n < (1 + 1/n)^n$ and as {$(1 - 2/n)$} is monotonically increasing.
So $0 < \lim_{n \rightarrow \infty} (1 - 2/n)^n < \lim_{n \rightarrow \infty} (1 + 1/n)^n$ so it exists.
You never said anything about wanting to know what it was, just that it exists.
A: All you have to do is use monotone convergence theorem, first show it's monotonic increasing or decreasing, by test Xn+1-Xn or Xn+1/Xn. 
Then show its bounded, at your case the smallest value occur at n=1, therefore it's bounded below by -1. 
Hope it helps
A: It is much easier to prove in general that the sequence $$f_{n}(x) = \left(1 + \frac{x}{n}\right)^{n}$$ tends to a limit as $n \to \infty$ for every real number $x$ and hence the limit $\lim_{n \to \infty}f_{n}(x)$ defines a new function $f(x)$.
Clearly when $x = 0$ then $f_{n}(x) = 1$ for all $n$ and hence the limit is $1$ so that $f(0) = 1$. Let's first start with the case $x > 0$. Clearly we can see via binomial theorem for positive integral index that $$f_{n}(x) = 1 + x + \dfrac{1 - \dfrac{1}{n}}{2!}\cdot x^{2} + \dfrac{\left(1 - \dfrac{1}{n}\right)\left(1 - \dfrac{2}{n}\right)}{3!}\cdot x^{3} + \cdots\tag{1}$$ where the sum on right is a finite sum with total of $(n + 1)$ terms. As $n$ increases each term increases and also the number of terms increases and each term is positive. It follows that $f_{n}(x)$ increases as $n$ increases. Moreover equation $(1)$ also shows that $$f_{n}(x) \leq 1 + x + \frac{x^{2}}{2!} + \cdots + \frac{x^{n}}{n!}$$ The series on right above converges (say to $E(x)$) as $n \to \infty$ and each term of the series is positive and hence it follows the partial sums of the series do not exceed the sum $E(x)$. It follows that $f_{n}(x) \leq E(x)$ for all $x$ where $$E(x) = 1 + x + \frac{x^{2}}{2!} + \cdots = \sum_{n = 0}^{\infty}\frac{x^{n}}{n!}$$ It follows that for each real number $x > 0$ the sequence $f_{n}(x)$ is increasing and bounded and hence the limit $f(x) = \lim_{n \to \infty}f_{n}(x)$ exists.

For $x < 0$ we need to consider a related sequence $$g_{n}(x) = \left(1 - \frac{x}{n}\right)^{-n}$$ and we can see that $$g_{n}(x) = \frac{1}{f_{n}(-x)}$$ From the above equation it is clearly seen that the convergence of $f_{n}(x)$ for $x < 0$ is solely dependent on the convergence of $g_{n}(x)$ for $x > 0$. Further if $g_{n}(x)$ converges to a non-zero value then $f_{n}(-x)$ also converges to a non-zero value. We will show that for $x > 0$ the sequence $g_{n}(x)$ converges to a non-zero value and this will ensure that $f_{n}(-x)$ also converges to a non-zero value. Thus the sequence $f_{n}(x)$ will be convergent for all values of $x$.
Clearly again via the general binomial theorem (for any index) we have $$g_{n}(x) = 1 + x + \dfrac{1 + \dfrac{1}{n}}{2!}\cdot x^{2} + \dfrac{\left(1 + \dfrac{1}{n}\right)\left(1 + \dfrac{2}{n}\right)}{3!}\cdot x^{3} + \cdots\tag{2}$$ which is valid for $0 < x < n$. The series on the right is an infinite series and clearly we can see that as $n$ increases the terms of series decrease so that $g_{n}(x)$ is decreasing as $n$ increases (after a certain value of $n$, namely after $n > x$). Moreover $g_{n}(x) \geq 1 + x$ and hence it follows that $g_{n}(x)$ converges as $n \to \infty$ for $x > 0$. Moreover the limit of $g_{n}(x)$ will not be less than $1 + x$ and hence will be positive.
The current question deals with the convergence of $(1 - 2/n)^{n} = f_{n}(-2)$ which is dependent on the convergence of $g_{n}(2)$ and we could have directly given the proof based on equation $(2)$.
A: $$\lim_{n\to \infty}\left(1-\frac{2}{n}\right)^n=\lim_{n\to \infty}\frac{1}{\left(\left(1+\frac{1}{\frac{n-2}{2}}\right)^{\frac{n-2}{2}}\right)^{\frac{2n}{n-2}}}$$
$$=\frac{1}{e^{\lim_{n\to \infty}\frac{2n}{n-2}}}=\frac{1}{e^{\lim_{n\to \infty}\left(2+\frac{4}{n-2}\right)}}=e^{-2}$$
