Single digit primes are different............and inside different sequences of 10 and powers thereof......are more clues.........from 10 - 20.......there are 4 primes............11, 13, 17, 19..........after single digit primes.............a prime number can ONLY end in a 1, 3, 7 or 9 (after single digit primes, i.e. primes greater than 10)..........no matter how big it is........................
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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