Prove that $\;\lim\limits_{n\to \infty}\left (1+\frac{1}{2^n}\right)^{2^n}$ exists I have to prove that the following limit exists:
$$\lim_{n\to \infty} \left(1+\frac{1}{2^n}\right)^{2^n}$$
I already proved it is strictly increasing, but I also have to prove its bounded. I need help with proving it to be bounded.
 A: Putting $2^n=r$ , so $r\to \infty$ as $n\to \infty$
So, the limit becomes $\;\lim\limits_{r\to \infty}\left(1+\dfrac 1 r\right)^r$
Now, its $(s+1)$th term of in binomial expansion is $$\frac{r(r-1)\cdots(r-s)}{r^s(s!)}=\frac 1{s!}\prod_{0\le t\le s}\left(1-\frac t r\right)$$
As $r\to\infty, (s+1)$ th term becomes  $\dfrac 1{s!}$
$$\lim_{r\to \infty}\left(1+\frac 1 r\right)^r=\sum_{0\le s< \infty}\frac 1{s!}=1+\frac 1 {1!}+\frac 1 {2!}+\cdots>2$$
Now, $3!=1\cdot2\cdot3>1\cdot2\cdot2\implies \dfrac 1{3!}<\dfrac 1{2^2}$
Similarly, $\dfrac 1{4!}<\dfrac 1{2^3}, \dfrac 1{5!}<\dfrac 1{2^4}$
So,
$\begin{align}\sum\limits_{0\le s<\infty}\dfrac 1{s!}<&1+1+\dfrac 1{2}+\dfrac 1{2^3}+\dfrac 1{2^4}+\cdots=\\=&1+\dfrac{1}{1-\dfrac 1 2}=1+2=3\end{align}$
So, $$2<\lim_{r\to \infty}\left(1+\frac 1 r\right)^r<3$$
A: To prove it is bounded, we will exploit the fact $ \ln(1+x) \leq x $. Now, we have
$$ a_n = e^{2^n\ln\left(1+\frac{1}{2^n}\right)} \leq e^{}  $$
A: Let's make two steps together, hopefully that should be enough.
Consider $f(x) = (1+1/x)^x$ and it is now sufficient to show that $f(x)$ is bounded as $x \to \infty$.
Let $L(x) = \log f(x) = x \log(1+1/x)$. Now since $\log(x)$ is continuous, it suffices to show that $L(x)$ is bounded as $x \to \infty$. (You could e.g. prove that $L(x) < 3$ for all $x > 1$.)
