Divergence test for $\sum_{n=1}^{\infty}\ln (1+\frac{1}{n})^n$. I am trying to prove that this is divergent 
$$\sum_{n=1}^{\infty} \left(1+\dfrac{1}{n}\right)^n$$
by finding the limit of 
$$\ln \left(1+\dfrac{1}{n}\right)^n$$
I know its $e$ and I am trying to arrive at that value by this
$$\ln y = n \ln(1 + \dfrac{1}{n})\\= \dfrac{\ln(1 + \dfrac{1}{n})}{\dfrac{1}{n}}$$
and I am already lost at this indeterminate form.

Let $f(x) = \ln(x)$.  Then using the definition of the derivative, we have:
\begin{align*}
\ln\left(\lim_{n\to \infty} \left(1+\frac{1}{n}\right)^n\right)&= \lim_{n\to \infty} \ln\left(\left(1+\frac{1}{n}\right)^n\right)
\\ &= \lim_{n\to\infty} n\ln\left(1+\frac{1}{n}\right)\\
&= \lim_{n\to\infty} \frac{\ln\left(1+\frac{1}{n}\right) - \ln(1)}{\frac{1}{n}}\\
&= f'(1)\\
&= 1
\end{align*}
Alternatively, using L'Hopital's Rule, we have:
\begin{align*}
\ln\left(\lim_{n\to \infty} \left(1+\frac{1}{n}\right)^n\right)&= \lim_{n\to \infty} \ln\left(\left(1+\frac{1}{n}\right)^n\right)
\\ &= \lim_{n\to\infty} n\ln\left(1+\frac{1}{n}\right)\\
&= \lim_{n\to\infty} \frac{\ln\left(1+\frac{1}{n}\right)}{\frac{1}{n}}\\
&= \lim_{n\to\infty} \frac{\frac{-1}{n^2+n}}{\frac{-1}{n^2}}\\
&= \lim_{n\to\infty} \frac{n^2}{n^2+n}\\
&= 1
\end{align*}
Note that being more rigorous with the computation of the limit in the above case (i.e., if you don't know L'Hopital's Rule or the computation of the derivative of the natural logarithm) is quite a bit more involved.  There is a third way to prove this which is even more advanced, which is to use the power series expansion of $\ln(1+x)$ about $x = 0$.  But all of these depend on either knowing the derivative of $\ln(x)$ or more properties of the logarithm.  But assuming these issues are resolved, we have:
$$\lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n = e$$
