Question about showing the harmonic series is divergent. Is it possible to show that the harmonic series is divergent by showing that the sequence of partial sums is a monotone increasing sequence that is unbounded?
 A: Perhaps you have seen the proof by the integral test, and you would like a more "hands on" proof, where you get a bound you can feel.  Here is such a proof:
$$
\begin{align}1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}...+\frac{1}{2^{n}-1} &> \frac{1}{2}+\left(\frac{1}{4}+\frac{1}{4}\right)+\left(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\right) ...+\frac{1}{2^n}\\
&=\frac{1}{2}+\frac{1}{2}+...+\frac{1}{2}\\
&=\frac{n}{2}
\end{align}
$$
So the sum of the first $2^n-1$ terms of the harmonic series is greater than $\frac{n}{2}$.  Since all the terms are positive, the sequence of partial sums is monotone increasing, and this special subsequence diverges to positive infinity, so we are done!  
A: Another way,
which is moderately equivalent.
Let
$S_n = \sum_{k=1}^n \frac1{k}$.
$S_n$ is obviously an increasing sequence.
$S_{2n}-S_n
=\sum_{k=1}^{2n} \frac1{k}-\sum_{k=1}^n \frac1{k}
=\sum_{k=n+1}^{2n} \frac1{k}
>\sum_{k=n+1}^{2n} \frac1{2n}
=\frac{n}{2n}
=\frac12
$
so, by induction
$S_{2^mn}-S_n
>\frac{m}{2}
$
so $S_{2^mn}$ can be made
as large as you want
by making $m$ large enough.
Note that this also shows that
$S_{2n}-S_n
=\sum_{k=n+1}^{2n} \frac1{k}
<\sum_{k=n+1}^{2n} \frac1{n}
=\frac{n}{n}
=1
$
so, by induction
$S_{2^mn}-S_n
<m
$
so to make $S_{2^mn}$
large,
you have to  go up exponentially.
This is why $S_n$ is about
$\log n$.
