Editing Harmonic series (mathematics) (section)

===Comparison test===
[[File:visual_proof_harmonic_series_diverges.svg|thumb|There are infinite blue rectangles each with area 1/2, yet their total area is exceeded by that of the grey bars denoting the harmonic series]]
One way to prove divergence is to compare the harmonic series with another divergent series, where each denominator is replaced with the next-largest [[power of two]]:
<math display=block>\begin{alignat}{8}
   1 & + \frac{1}{2} && + \frac{1}{3} && + \frac{1}{4} && + \frac{1}{5} && + \frac{1}{6} && + \frac{1}{7} && + \frac{1}{8} && + \frac{1}{9} && + \cdots \\[5pt]

   {} \geq 1 & + \frac{1}{2} && + \frac{1}{\color{red}{\mathbf{4}}} && + \frac{1}{4} && + \frac{1}{\color{red}{\mathbf{8}}} && + \frac{1}{\color{red}{\mathbf{8}}} && + \frac{1}{\color{red}{\mathbf{8}}} && + \frac{1}{8} && + \frac{1}{\color{red}{\mathbf{16}}} && + \cdots \\[5pt]
 \end{alignat}</math>
Grouping equal terms shows that the second series diverges (because every grouping of convergent series is only convergent):
<math display=block>\begin{align}
   & 1 + \left(\frac{1}{2}\right) + \left(\frac{1}{4} + \frac{1}{4}\right) + \left(\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}\right) + \left(\frac{1}{16} + \cdots + \frac{1}{16}\right) + \cdots \\[5pt]
   {} = {} & 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \cdots.
 \end{align}</math>
Because each term of the harmonic series is greater than or equal to the corresponding term of the second series (and the terms are all positive), and since the second series diverges, it follows (by the [[Direct comparison test|comparison test]]) that the harmonic series diverges as well.  The same argument proves more strongly that, for every [[Positive number|positive]] {{nowrap|[[integer]] <math>k</math>,}}
<math display=block>\sum_{n=1}^{2^k} \frac{1}{n} \geq 1 + \frac{k}{2}</math>
This is the original proof given by [[Nicole Oresme]] in around 1350.{{r|kifowit}} The [[Cauchy condensation test]] is a generalization of this argument.{{r|roy}}