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