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 $1$. 

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 $e$. 
