Prove that \(\lim\big(n)\) diverges.

For the definitions of sequences and what it means to for a sequence to converge, see this, for subsequences see this.
For the “show the limit” template, see this.

Problem Discussion

From the definitions page, a sequence does not converge to \(L\) if there exists some \(\epsilon\) such that for all tails of the sequence, the tail will not be inside the interval \((L-\epsilon, L+\epsilon)\). In other words, for all \(N \in \mathbb{N}\), there is some \(n > N\) where \(n\) is not inside the interval \((L-\epsilon, L+\epsilon)\).

The sequence diverges if this is true for any \(L\). So we need to show that \(\lim\big(n)\) doesn’t converge for any possible \(L\).

Formal Proof

Suppose for the sake of contradiction that the sequence does converge for some \(L \in \mathbb{R}\). We will show that there exists some \(\epsilon > 0\) such that for all \(N \in \mathbb{N}\), there exists an \(n > N\) with

$$ \begin{align*} n \not\in (L-\epsilon, L+\epsilon) \end{align*} $$

Let \(\epsilon = 1\), then the interval becomes \((L-1, L+1)\). Observe that \({n}^{\infty}_{n=1}\) is unbounded so we can always find an \(n > N\), such that \(n > L + 1\) which implies \(n \not\in (L-1, L+1)\). This contradicts the definition of convergence. Thus, the sequence \({n}^{\infty}_{n=1}\) can not converge to \(L\). Therefore, the sequence diverges as we wanted to show. \(\blacksquare\)