# Is \sum_{i=1}^n 1/n divergent?

**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.