finding whether this sequence converges Let
$$a_n=\left(1+\frac{1}{n}\right)^n$$
Does this converge?
When $n$ goes to $\infty$ we end up with $1$ to the power of $\infty$. What does this mean to us? Is it $1$ or undefined. I remember from calculus course that this is not defined, so we take it as divergent right? But why diverges? $1$ to the infinity is $1$. 
 A: When I first learned about this sequence it was pretty counterintuitive to me that it converges (for the same reasons you mentioned). Actually you have that $$a_n=(1+\frac{1}{n})^n \to e$$
for $n \to \infty$. You can show that by noticing that the sequence is increasing and bounded.
$(a_n)_{n \in \mathbb N}$ is increasing:
We need to use the Bernoulli inequaltiy: For all $x \in \mathbb R$ with $1+x > 0$ and $x\neq 0$ and all $n \geq 2$ we have $1 + nx < (1+x)^n$ (Bernoulli). Hence using the Bernoulli inequaltiy we get
$$ (1+\frac 1 n)^n (1-\frac 1 n)^n = (1-\frac{1}{n^2})^n > 1 - \frac 1 n.$$
This is equivalent to
$$ (1+\frac 1 n)^n (1-\frac 1 n)^{n-1} > 1.$$
Hence we get for all $n \geq 2$ that
$$ a_n = (1+\frac 1 n)^n > \frac{1}{(1-\frac 1 n)^{n-1}} = (\frac{n}{n-1})^{n-1} = (1 + \frac{1}{n-1})^{n-1} = a_{n-1}.$$
$(a_n)_{n \in \mathbb N}$ is bounded: It is clear that $a_n \geq 1$ for all $n \in \mathbb N$. Now consider $b_n = (1+ \frac 1 n)^{n+1}$. We will show that $(b_n)_{n \in \mathbb N}$ is decreasing. Using the Bernoulli inequality we achieve
$$ \frac{1}{(1+\frac 1 n)^n (1-\frac 1 n)^n} = \frac{1}{(1-\frac{1}{n^2})^n} = (1 + \frac{1}{n^2 - 1})^n > (1 + \frac{1}{n^2})^n > 1 + \frac{1}{n}.$$
Hence we get for all $n \geq 2$ that
$$b_{n-1} = (1 + \frac{1}{n-1})^n = (\frac{n}{n-1})^n > (1+\frac 1 n)^{n+1} = b_n.$$
Since $(b_n)_{n \in \mathbb N}$ is decreasing we have that
$$4 = b_1 \geq b_n = (1+ \frac 1 n)^{n+1} \geq (1+ \frac 1 n)^{n} = a_n$$
for all $n \in \mathbb N$. Hence $(a_n)_{n \in \mathbb N}$ is bounded.
Because $(a_n)_{n \in \mathbb N}$ is increasing and bounded it converges and its limit is called $e$ in literature.
I hope it helps you :)
A: We first prove that the sequence $a_n=\left(1+\frac1n\right)^n$ converges.  To do so, we will show that it is monotonically increasing and bounded above.  The Monotone Convergence Theorem guarantees that such a sequence converges.


SHOWING THAT $a_n$ MONOTONICALLY INCREASES

To see that $a_n$ is monotonically increasing, we analyze the ratio $\frac{a_{n+1}}{a_n}$.  Proceeding, we find
$$\begin{align}
\frac{a_{n+1}}{a_n}&=\frac{\left(1+\frac1{n+1}\right)^{n+1}}{\left(1+\frac1n\right)^n}\\\\
&=\left(1+\frac1n\right)\left(\frac{n(n+2)}{(n+1)^2}\right)^{n+1}\\\\
&=\left(1+\frac1n\right)\left(1-\frac{1}{(n+1)^2}\right)^{n+1} \tag 1\\\\
&\ge \left(1+\frac1n\right)\left(1-\frac{1}{(n+1)}\right) \tag2\\\\
&=1
\end{align}$$ 
where we applied Bernoulli's Inequality in going from $(1)$ to $(2)$.  

Therefore, $a_n$ is monotonically increasing.



SHOWING THAT $a_n$ IS BOUNDED ABOVE

Using the Binomial Theorem, we have
$$\begin{align}
a_n&=\left(1+\frac1n\right)^n\\\\
&=1+1+\frac{1}{2!}\frac{n(n-1)}{n^2}+\frac{1}{3!}\frac{n(n-1)(n-2)}{n^3}+\cdots +\frac{1}{n!} \tag 3\\\\
&\le 1+1+\frac{1}{2!}+\frac{1}{3!}+\cdots +\frac1{n!}\tag 4\\\\
&\le 1+1+\frac12+\frac14+\cdots +\frac1{2^n}\tag 5\\\\
&\le 1+\sum_{k=0}^n \frac{1}{2^k}\tag 6\\\\
&=1+\frac{1-2^{n+1}}{1-1/2}\tag 7\\\\
&\le 3
\end{align}$$
In going from $(3)$ to $(4)$, we observed that $\frac{n(n-1)(n-2)\cdots (n-k))}{n^{k+1}}\le 1$ for all $k\ge1$.
In going from $(4)$ to $(5)$, we noted that $k!\ge 2^{k-1}$ for $k\ge 1$.
In going from $(5)$ to $(6)$, we simply wrote the sum using the summation notation.
In going from $(6)$ to $(7)$, we summed a Geometric Progression.

Therefore, $a_n$ is bounded above (by $3$).


Since $a_n$ is monotonically increasing and bounded above, the monotone convergence theorem guarantees that the $a_n$ is a convergent sequence.

NOTE:
The limit of the sequence $a_n$ is one of the Representations of Euler's number $e$. 
A: If we define the logarithm function as:
$$\ln x=\int_1^x\frac{1}{t}dt$$
we can easily verify that $\ln x$ is analytic at $x=1$ and has a Maclaurin series in its neiborhood:
$$\ln(1+t)=\sum_{i=1}^\infty (-1)^{i+1}\frac{t^i}{i},\;\;|t|<1$$
The rest would be straightforward:
$$\ln a_n=\ln\left(1+\frac{1}{n}\right)^n=n\sum_{i=1}^\infty\frac{(-1)^{i+1}}{i}\left(\frac{1}{n}\right)^i
=n\left(
\frac{1}{n}-\frac{1/n^2}{2}+\frac{1/n^3}{3}-\cdots
\right)\\
=1-\frac{1}{2n}+\frac{1}{3n^2}-\cdots$$
Taking the limit as $n\to\infty$ results in $\lim_{n\to\infty}\ln a_n=1$.
