I need help figuring out how this sequence converges. $A_n = ( 1 + \frac  2n ) ^ n$;
I know the end result is convergent at $e^2$, but how do I figure that out? I've started setting it up as $An = (1 + (2/n))^{(n/2)2}$ as someone has suggested but i don't understand how that results in $e^2$.
 A: If you know that $(1 + \frac 1n)^n$ tends to $e$ as $n\to\infty$, then you can write down as your friends suggested by taking $n'=\frac n2$:
$$
\lim_{n\to\infty}(1 + \frac 2n)^n=\lim_{n\to\infty}\left(1 + \frac 1{\frac n2}\right)^{2(\frac n2)}=\lim_{n'\to\infty}\left((1 + \frac 1{n'})^{n'}\right)^2=e^2
$$
A: The suggestion that you received relies on your knowing that
$$\lim_{x\to 0^+}(1+x)^{1/x}=e\;.$$
Given that knowledge, you can substitute $x=\frac2n$ and compute
$$\begin{align*}
\lim_{n\to\infty}\left(1+\frac2n\right)^n&\lim_{x\to 0^+}(1+x)^{2/x}\\\\
&=\lim_{x\to 0^+}\left((1+x)^{1/x}\right)^2\\\\
&=\left(\lim_{x\to 0^+}(1+x)^{1/x}\right)^2\\\\
&=e^2\;.
\end{align*}$$
With a little practice one needn’t actually make the substitution:
$$\begin{align*}
\lim_{n\to\infty}\left(1+\frac2n\right)^n&=\lim_{n\to\infty}\left(1+\frac2n\right)^{(n/2)2}\\\\
&=\lim_{n\to\infty}\left(\left(1+\frac2n\right)^{n/2}\right)^2\\\\
&=\left(\lim_{n\to\infty}\left(1+\frac2n\right)^{n/2}\right)^2\\\\
&=e^2\;.
\end{align*}$$
A: $$ \lim_{n \to \infty} \left(1 + \frac{2}{n} \right)^n = e^{\lim_{n \to \infty} \frac{\ln \left( 1 + \frac{2}{n} \right) }{\frac{1}{n}}} \overset{\ast}{=} e^{\lim_{n\to\infty } \frac{-\frac{2}{n^2}\frac{1}{1 + 2/n}}{-1/n^2}} = e^{\lim_{n\to \infty} \frac{2}{1+2/n} } = e^2  $$
Where in $ \ast $ we used L'HÃ´pital's Rule
