Show that the sequence $ln(n)$ diverges I am trying to show that $\ln(n)$ diverges to $\infty$. Here is my work. 
By way of contradiction, suppose that $\ln (n)$ converges to $\ell$; i.e., $\lim_{n \rightarrow \infty} \ln(n) = \ell$. Since the natural log function is surjective, there exists $y$ such that $\ln (y) = \ell$, giving us $\lim_{n \rightarrow \infty} \ln (n) = \ln(y)$, and therefore $\lim_{n \rightarrow \infty} (\ln(n) - \ln(y)) = 0$, or finally $\lim_{n \rightarrow \infty} \ln(n/y) = 0$. Letting $k = \frac{n}{k}$, then it seems $\lim_{k \rightarrow \infty} \ln(k) = 0$. Now, consider $\epsilon = \ln(2) > 0$. Then there exists $N \in \mathbb{N}$ such that $\ln(k) < \ln(2)$ for every $k \ge N \ge 1$. Hence, $k < 2$ and $k=2$ produces one of many contradictions.
Is this correct? I do have one general question relating to this that is somewhat subtle in nature: how do I formally justify the substitution? Intuitively, I realize that if $k = \frac{n}{y}$, then as $n \rightarrow \infty$, $k \rightarrow \infty$, and so $\lim_{n \rightarrow \infty} \ln (\frac{n}{y}) = \lim_{k \rightarrow \infty} \ln (k)$, but I am having difficulty seeing how one formally justifies this. 
 A: I would use a proof by contradiction as well, but slightly different in order to hit at the heart of the problem: 
Definition of a limit going to infinity:
We say that $\lim_{n\rightarrow \infty}f(n) =L$ iff for every $\epsilon > 0$, there exists an $N$ such that $|f(n)-L| < \epsilon$ when $n>N$. 
Suppose, for sake of contradiction, that $\lim_{n\rightarrow \infty}\ln(n) \neq \infty$, that $\lim_{n\rightarrow \infty}\ln(n) = L$, where $L$ is a finite number. We use the definition of a limit going to infinity to say that if we set $\epsilon$ to $0.01$, there must be a number $N$ such that $|\ln(n)-L| < 0.01$ when $n>N$. 
Therefore, we have if $n>N$: 
$$-0.01<\ln(n)-L<0.01$$
$$-0.01+L<\ln(n)<0.01+L$$
$$e^{-0.01+L} < n < e^{0.01+L}$$
$$e^{-0.01} \cdot e^L < n <e^{0.01} \cdot e^L $$
Since $e^{0.01}$, $e^{-0.01}$, and $e^L$ are all constants, we have, for all $n>N$: $CD<n<CE$, where $CD$ and $CE$ are finite. Therefore, we now say that the $\lim_{n\rightarrow \infty}n=L$, where $L$ is finite and between $CD$ and $CE$. 
This is a contradiction, because $\lim_{n\rightarrow \infty}n=\infty$. Therefore, $\lim_{n\rightarrow \infty}\ln(n) \neq L$, where $L$ is finite. Therefore,  $\lim_{n\rightarrow \infty}\ln(n) = \infty$, and $\ln(n)$ diverges. 
Q.E.D
Perhaps this is slightly more rigorous that you were looking for, but you cannot simply say that you can see that the derivative of $\ln(x)$ is always positive, because other functions that do converge (like $-1/x$) fulfill this requirement. 
A: A different way: since $\exp$ is strictly increasing and goes to infinity, $\log(n) \to \infty$ iff $\exp(\log(n))$ does.
A: By definitions and the particular Rieman sum with unit length intervals to the right of $x=2,$
$$\displaystyle\ln n = \int _1^n\frac{1}{t}dt \geq \sum_{i=2}^n \frac{1}{i}=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n},$$
which is famously divergent!
A: The limit game goes like this:
Pick a really big number and call it $N$.
Find a value for $n$ such that $\ln{n}\gt N$.
If you can do this no matter how big $N$ is, then $\displaystyle{\lim_{n\to\infty}\ln{n}=\infty}$.
Edit: you need the fact that $\ln{n}$ is monotone increasing to properly use this argument.
A: Clearly, the main question is - as usual : "how a mathematical object is defined ?"
I will consider that $\ln$ denotes the primitive of $f:(0,+\infty)\to\mathbb{R}$ which vanishes at $x=1$.
From that starting point, it's easy to prove the "functional equation for the logarithm", that is :
$$\forall (x,y)\in(0,+\infty)^2,\,\ln(xy)=\ln(x)+\ln(y)$$
Since $\ln$ is strictly increasing, there are two possibilities regarding its behavior at $+\infty$ :


*

*either $\lim_{x\to+\infty}\ln(x)=L\in\mathbb{R}$

*or $\lim_{x\to+\infty}\ln(x)=+\infty$


Now suppose that the first case holds. We know that, for all $x>0$ : 
$$\ln(2x)=\ln(x)+\ln(2)$$
Taking the limits in both sides leads to : $L=L+\ln(2)$, a contradiction (indeed, $\ln(2)>\ln(1)=0$).
Hence, all this proves that:

$$\lim_{x\to+\infty}\ln(x)=+\infty$$

And, as a consequence, the sequence $(\ln(n))_{n\ge1}$ diverges to $+\infty$, just as any sequence of the form $(\ln(t_n))_{n\ge1}$ where $(\forall n\ge1,t_n>0)$ and $\lim_{n\to\infty}t_n=+\infty$.
