Editing Harmonic series (mathematics) (section)

==Definition and divergence{{anchor|divergence}}==
The harmonic series is the infinite series
<math display=block>\sum_{n=1}^\infty\frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \cdots</math>
in which the terms are all of the positive [[unit fraction]]s. It is a [[divergent series]]: as more terms of the series are included in [[partial sum]]s of the series, the values of these partial sums grow arbitrarily large, beyond any finite limit. Because it is a divergent series, it should be interpreted as a formal sum, an abstract mathematical expression combining the unit fractions, rather than as something that can be evaluated to a numeric value. There are many different proofs of the divergence of the harmonic series, surveyed in a 2006 paper by S. J. Kifowit and T. A. Stamps.{{r|kifowit}}
Two of the best-known{{r|rice|kifowit}} are listed below.

===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}}

===Integral test===
[[File:Integral Test.svg|thumb|Rectangles with area given by the harmonic series, and the hyperbola <math>y=1/x</math> through the upper left corners of these rectangles]]
It is possible to prove that the harmonic series diverges by comparing its sum with an [[improper integral]].  Specifically, consider the arrangement of rectangles shown in the figure to the right. Each rectangle is 1 unit wide and <math>\tfrac1n</math> units high, so if the harmonic series converged then the total area of the rectangles would be the sum of the harmonic series. The curve <math>y=\tfrac1x</math> stays entirely below the upper boundary of the rectangles, so the area under the curve (in the range of <math>x</math> from one to infinity that is covered by rectangles) would be less than the area of the union of the rectangles. However, the area under the curve is given by a divergent [[improper integral]],
<math display=block>\int_1^\infty\frac{1}{x}\,dx = \infty.</math>
Because this integral does not converge, the sum cannot converge either.{{r|kifowit}}

In the figure to the right, shifting each rectangle to the left by 1 unit, would produce a sequence of rectangles whose boundary lies below the curve rather than above it.
This shows that the partial sums of the harmonic series differ from the integral by an amount that is bounded above and below by the unit area of the first rectangle:
<math display=block>\int_1^{N+1}\frac1x\,dx<\sum_{i=1}^N\frac1i<\int_1^{N}\frac1x\,dx+1.</math>
Generalizing this argument, any infinite sum of values of a monotone decreasing positive function {{nowrap|of <math>n</math>}} (like the harmonic series) has partial sums that are within a bounded distance of the values of the corresponding integrals. Therefore, the sum converges if and only if the integral over the same range of the same function converges. When this equivalence is used to check the convergence of a sum by replacing it with an easier integral, it is known as the [[integral test for convergence]].{{r|bressoud}}