Alternative ways to show that the Harmonic series diverges

This question came to me from one of my calculus students today: Other than using the integral test $$\int_1^\infty \frac{dx}{x} \to \infty,$$ what are some other ways that we can prove the Harmonic series $\sum_{n=1}^\infty \frac{1}{n}$ diverges?

I’m sure there are plenty of methods out there; any method where a typical student in Calculus could understand would be great.

Solutions Collecting From Web of "Alternative ways to show that the Harmonic series diverges"

Try applying the Comparison test to the harmonic series, specifically to this series

$1 + 1/2 + 1/4 + 1/4 + 1/8 + 1/8 + 1/8 + 1/8 + 1/16 + …$

= $1 + 1/2 + (1/4+1/4) + (1/8+1/8+1/8+1/8) + …$

= $1+ 1/2 + 1/2 + 1/2 + … = \infty$

Since each term is larger than the term in the above series, the harmonic series diverges as well.

The “easiest” way in my opinion is to remark that

$$\sum_{k=n+1}^{2n} \frac{1}{k} \geq n\frac{1}{2n} = \frac{1}{2}$$

So if we cut the sum between $1$ and $2^n$ by powers of two, we have

$$\sum_{k=1}^{2^n} \frac{1}{n} = \sum_{k=1}^{n} \sum_{i=2^{k-1}+1}^{2^k} \frac{1}{i} \geq \sum_{k=1}^n \frac{1}{2} = \frac{n}{2}$$

Consider the generating function
$$f(x) = \sum_{k=1}^\infty \frac{x^n}{n}$$
and differentiate in the radius of convergence to get
$$f'(x) = \frac{d}{dx} \sum_{k=1}^\infty \frac{x^n}{n} = \sum_{k=1}^\infty x^{n-1} = \frac{1}{1-x},$$
which only converges for $|x| < 1$, so $f(x)$ will only be defined in the same interval, and the original series is $f(1)$, which diverges.

Suppose

$$S=1+{1\over2}+{1\over3}+{1\over4}+\cdots$$

is convergent. Since all the terms are non-negative, it must be absolutely convergent. But absolutely convergent sequences can be fiddled with at will, allowing us to conclude

\begin{align} 2S&=2+1+{2\over3}+{1\over2}+\cdots\\ &=S+\left(2+{2\over3}+{2\over5}+\cdots\right)\\ &\gt S+\left(1+{1\over2}+{1\over3}+\cdots\right)\\ &=2S \end{align}

which is absurd: No number is greater than itself.