I was studying the definition of limits of a sequence in $\mathbb R$ and tried to verify whether the sequence $X=(1,2,3,......)$ converges to a real number or not.
Solution: Here, the series does not seem to converge to a certain value. To support this precursory
intuition, I must show that no real number can be the limit of this sequence. Let $x\in \mathbb R$.
Let us choose $\epsilon=1$. For any $K(\epsilon)\in \mathbb N$, we need to show that for some $n\in \mathbb N$ with $n\geq K(\epsilon)$, $|x_n-x|\geq \epsilon=1$. We can find a $h$ in $\mathbb N$ such that $h>x$ (The Archimedean property of real numbers).If $K(\epsilon)\leq h$, then choosing $n=h+1$ we get, $$|x_n-x|=|h+1-x|=|h-x+1|=h-x+1>1$$
If $h<K(\epsilon)$, then choose $n=K(\epsilon)+1$ to get $$|x_n-x|=|K(\epsilon)+1-x|>1$$This is because $x<h<K(\epsilon)\Rightarrow 0<K(\epsilon)-x \Rightarrow 1<K(\epsilon)-x+1$.
Let me know if my proof is correct. Further, since I am new to the concept of convergence, though I am able to understand the definition, I want some intuition and 'big picture' on the concept of limits of sequence.
Well you can use the theorem that convergent sequence is always bounded, that means if $(x_n)$ be a convergent real sequence, then $\exists M>0$ such that $|x_n|< M\hspace{10pt}\forall n\in\mathbb{N}$.
Given $x_n=n\hspace{10pt}\forall n\in\mathbb{N}$. Suppose if possible, there exists a real number $M>0$ such that $|x_n|=x_n<M$ for all $n\in\mathbb{N}$. By Archimedean property we know that the set $\mathbb{N}$ is unbounded above, i.e there does not exists any such $M>0$ for which the above condition holds, which contradicts our assumption that $(x_n)=(n)$ is bounded.
Hence we conclude, the sequence is not convergent.