am I allowed to choose epsilon in this way in cauchy sequence?

Question

Its given that $a_{n+1}-a_{n} >= \frac{1}{n}$ We need to show that $a_{n}$ goes for infinity, My solution: from the given we conclude $a_{n}$ is monotonic,lets assume $a_{n}$ is bounded then it converges to $L$, a sequence converges iff its cauchy sequence thus it must satisfy that $\vert a_{n+p}-a_{n}\vert \leq \epsilon $

For $p=1$, $\epsilon=\frac{1}{n}$ we get that it cant be a cauchy sequence according to the first given Thus it's not bounded thus it goes to infinity

My question is: am I allowed to choose epsilon to be $\frac{1}{n}$, is my solution valid?

Thanks

Answer 1

You cannot have a new epsilon for every new n when n goes to infinity.

$a_{n+1}>a_n+1/n >a_{n-1}+1/(n-1)+1/n>…>a_1+H(n)$ where H(n) is the sum of the harmonic series. You might then need to prove that this series diverges. That proof is easier to find, e.g. on Wikipedia.

Answer 2

The $\epsilon$ we use is termed "arbitrary", that is, it does not depend on anything. Here your $\epsilon$ depends on $n$, so it is not feasible.

However for this proof, I can give you a hint: $$a_{n+1}-a_n \ge \frac{1}{n} \Rightarrow \sum_{n=1}^{\infty}(a_{n+1}-a_n) \ge \sum_{n=1}^{\infty} \frac{1}{n} $$

The R.H.S. of inequality is divergent. Can you use comparison test now?