Establish the convergence and find the limits of the following sequence $a_n = \left(1+\dfrac{1}{n}\right)^{n+1}$
I know that the answer is supposed to be $e$ but I am unsure how to reach that answer.
I am so lost where to even begin with this
 A: Hint: $a_n = \left(1+\dfrac{1}{n}\right)^n\cdot \left(1+\dfrac{1}{n}\right)$
A: Unfortunately I dont think there is any way to "prove" that:
$$ \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n = e$$
At least, not without circular reasoning.
The fundamental problem is that this limit IS THE definition of Euler's Number.  This is how Jacob Bernoulli defined it.
All other properties (the ones that might be used to prove this) stem from the truth of this definition.

Im fairly confident that if you took the binomial expansion:
$$ \left(1 + \frac{1}{n}\right)^n = \sum_{k=0}^n {n\choose k}\left(\frac{1}{n}\right)^k$$
Then took the limit as $n\to\infty$, using, say, the Ratio Test or something, you should be able to prove convergence.
Actually, it might be easier to use the monotone convergence criterion. Let $s_n=(1+\frac1n)^n$ and you can easily show with inductive arguments both that the sequence increases and is bounded above by 3.
Then, simply declare that the limit is called $e$, which is what Bernoulli did.
A: $$(1)\ \lim_n a_n = \lim _n e^{\ (n+1)\cdot\ln(1+\frac{1}{n})} = e^{\lim \limits_n \ (n+1) \cdot \ln(1+\frac{1}{n})} = e^1$$
$\textbf{Comment}$: Verfify by L'H (or however you see fit) that, $$\lim_{n} \frac{\ln \big(1+ \frac{1}{n}\big)}{\frac{1}{(n+1)}} = \lim_{x} \bigg(\frac{1}{x(x+1)}\bigg)(x+1)^2 = \lim_{x} \frac{x+1}{x}=1.$$
$\textbf{Recall}$: From calculus that if $f(x)$ is a continuous function and $(a_n) \to L$ then $f(a_n) \to f(L)$. In this case $f(x)=e^x$ is our continuous function. 
A: Hint:
$$a_n=\left(1+\frac{1}{n}\right)^{n+1}$$
$$\ln a_n=\ln \left(1+\frac{1}{n}\right)^{n+1}$$
$$\ln a_n=(n+1) \ln \left(1+\frac{1}{n}\right)$$
