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 $1$. 

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 $\lim_{n\to\infty}\ln a_n=1$.
