On this page, we prove that the harmonic series is a divergent series. In fact, this is one of the best-known and classic examples of the infinite series course.
Why does the harmonic series diverge
We start by proving classical results on sequences. If a sequence $(u_n)_n$ is increasing and has no upper bound. Then $\lim_{n\to+\infty}u_n=+\infty$. In fact, by assumption for any $M>0$, there exist $N\in \mathbb{N}$ such that $u_N>M$. Now by the fact that $(u_n)_n$ is increasing, for any $n>N$ we have $u_n\ge u_N>M$. This is exactly the definition of $\lim_{n\to \infty}u_n=+\infty$.
Now we shall use this result to prove that the following harmonic series satisfies $$ \sum_{n=1}^{+\infty} \frac{1}{n}=+\infty.$$ To this end, we select $$ H_n=1+\frac{1}{2}+\cdots+\frac{1}{n}=\sum^{n}_{k=1}\frac{1}{k},\qquad (n\ge 1).$$ Clearly, we have $H_{n+1}-H_n=\frac{1}{n}>0$. Thus $(H_n)_n$ is strictly increasing. According to the above result to prove that $\lim_{n\to+\infty}H_n=+\infty,$ it suffices to show that the sequence has no upper bound. By contradiction, assume that this sequence has an upper bound. Then it is a convergent sequence. Thus, there exists a real number $\ell$ such that $H_n \to \ell,$ as $n\to\infty$. So that $H_{2n}\to \ell$ as $n\to \infty,$ and then $H_{2n}-H_n\to 0$ as $n\to\infty$. On the other hand, we have $$ H_{2n}-H_n=\sum_{k=n+1}^{2n} \frac{1}{k}\ge \sum_{k=n+1}^{2n} \frac{1}{2n}=\frac{1}{2}.$$ Now by letting $n\to\infty,$ we obtain $0\ge \frac{1}{2},$ a contradiction.