Proof:

Consider the sequence $$\{a_n\}$$, where $$a_n = \sum_{i=1}^n 1/i$$.

$$a_n$$ is convergent if and only if $$\lim_{n \to \infty} a_n$$ exists.

Let, the limit is L. That is, $$\lim_{n \to \infty} a_n = L$$.

This implies that for any arbitrarily small $$\epsilon > 0$$, there exists a natural number N such that $$| a_n - L | < \epsilon$$ for all $$n \geq N$$.

[A] Now, consider $$\epsilon > 0$$ such that $$\epsilon < 1/2$$. Then there exists a natural number N such that $$| a_n - L | < \epsilon$$ for all $$n \geq N$$. Consider two such natural numbers $$m, n$$ such that $$m > n > N$$.

We can compute the difference between $$a_m$$ and $$a_n$$ as follows: $$|a_m - a_n|$$
$$= | \sum_{i=1}^m 1/i - \sum_{i=1}^n 1/i |$$
$$= | \sum_{i=n+1}^m 1/i | = (\frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{m})$$
$$> (m-n)\frac{1}{m}$$
$$= \frac{m-n}{m}$$
$$= 1 - \frac{n}{m}$$
$$\geq 2*\epsilon$$, since, choice of $$\epsilon$$ is arbitrary and we may choose a small enough $$\epsilon$$ to satisfy this.

This implies that $$| a_m - a_n | \geq 2\epsilon$$

Since, $$a_n \in (L-\epsilon, L+\epsilon)$$, this implies that $$a_m \notin (L-\epsilon, L+\epsilon)$$. This contradicts statement [A].

This implies that $$(a_n)$$ is not convergent.

The proof is complete.