The series from 2 to 11......has ten whole numbers in it.............but 11 - 2 = 9.........clues from Nature..................i am still very sure that there is a prime fund..........2, 3, 5, 7, 11....................the harmonic fundamental is 1..........the length of a string of a musical instrument................the fractions are the over tones.......of the length of the string.....
Comparison test[edit]
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:
![{\displaystyle {\begin{aligned}&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 \\[12pt]>\;\;\;&1\;\;+\;\;{\frac {1}{2}}\;\;+\;\;{\frac {1}{4}}\,+\,{\frac {1}{4}}\;\;+\;\;{\frac {1}{8}}\,+\,{\frac {1}{8}}\,+\,{\frac {1}{8}}\,+\,{\frac {1}{8}}\;\;+\;\;{\frac {1}{16}}\,+\,\cdots \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d33be1134374a704aed1d911722c4095554015c1)
Each term of the harmonic series is greater than or equal to the corresponding term of the second series, and therefore the sum of the harmonic series must be greater than the sum of the second series. However, the sum of the second series is infinite:
![{\displaystyle {\begin{aligned}&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 \\[12pt]=\;\;&1\;\;+\;\;{\frac {1}{2}}\;\;+\;\;{\frac {1}{2}}\;\;+\;\;{\frac {1}{2}}\;\;+\;\;{\frac {1}{2}}\;\;+\;\;\cdots \;\;=\;\;\infty \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a59ba63c85f07a3cfbb009696fc135d757be3052)
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