Let $a_n$ be a sequence such that: $ a_{n+1}-a_n \ge \frac{1}{n}$. prove that $\lim_\limits{n \to \infty} a_n = \infty$. Let $a_n$ be a sequence such that: $ a_{n+1}-a_n  \ge \frac{1}{n}$. prove that $\lim_\limits{n \to \infty} a_n = \infty$.
SOLUTION: I want to prove this by disproving Cauchy's test. which means, I want to prove that:
There is an $\epsilon>0$ , such that for each $N$, there is $n>N$ and $p \in$  $\mathbb N$ such that:$|a_{n+p}-a_n| \ge \epsilon$.
but for some reason I couldn't find that $\epsilon$. the "$\frac{1}{n}$" made my job harder. 
To conclude: I want to disprove that $a_n$ converges by disproving that its Cauchy's and from $ a_{n+1}-a_n  \ge \frac{1}{n}$ we get that the sequence is monotonic increasing, thus, this will prove that: $\lim_\limits{n \to \infty} a_n = \infty$.
Am I on the right track? are all the conclusions that I've reached are right? How do I really complete this? any kind of help would be appreciated. 
 A: Proof the sequence is not Cauchy
Using what you know about: $a_{n+1} - a_n \geq \frac{1}{n}$ you have:
$|a_{n+p} - a_{n}| = |\sum_{i=0}^{p-1}(a_{n+i+1} - a_{n+i})| \\
= \sum_{i=0}^{p-1}(a_{n+i+1} - a_{n+i}) \quad \text{because} \quad a_{n+1} - a_n \geq \frac{1}{n} \geq 0\\
\geq \sum_{i=0}^{p-1}\frac{1}{n+i} \\
\geq \sum_{i=0}^{p-1}\frac{1}{n+p}$
You now have:
$|a_{n+p} - a_{n}| \geq \frac{p}{n+p}$
You also have: $lim_{p \to +\infty}\frac{p}{n+p} = 1$
This implies that whatever $n$ you choose, you will find a $p$ such that $\frac{p}{n+p} > \frac{1}{2}$
So if you choose $\epsilon = \frac{1}{2}$ you have your contradiction for the Cauchy.
Proof a monotonicaly increasing sequence can converge
As you stated that:

the sequence is monotonic increasing, thus, this will prove that: $lim_{n \to +\infty}{a_n}=+\infty$

This is not true, a counter example is: $1-\frac{1}{n}$
Indeed, there is a result that monotonically increasing bounded sequences do converge. 
A: Hint : you should consider $$\lim_{N\to\infty} \sum_{n=0}^N (a_{n+1}-a_n)=\lim_{N\to\infty} a_{N+1}-a_0$$
Edit : since sum of series isn't in the material : 
$$\begin{align}
\lim_{N\to\infty} a_{N+1}-a_1 &=\lim_{N\to\infty} \sum_{n=1}^N (a_{n+1}-a_n) \\\\
&\ge \lim_{N\to\infty} \sum_{n=1}^N \frac{1}{n} \\\\
&\ge \lim_{N\to\infty} \sum_{n=1}^N \int_n^{n+1} \frac{1}{t}dt\\\\
&= \lim_{N\to\infty} \int_1^{N+1} \frac{1}{t}dt\\\\
&= \lim_{N\to\infty} \int_1^{N+1} \frac{1}{t}dt\\\\
&= \lim_{N\to\infty} \ln(N+1) = \infty
\end{align}$$
A: Hint: Look at a proof for the divergence of the harmonic series and apply to your situation by avoiding the word "series".
A: Hint Show that 
$$a_{2^{n+1}}-a_{2^n} \geq \frac{2^{n}}{2^{n+1}}$$
A: $$a_{n+1}-a_1=\sum_{k=1}^{n}a_{k+1}-a_k  \ge 1 + \frac{1}{2} +\frac{1}{3}++..++\frac{1}{n}\to \infty$$
