Every time there is a successive power of ten......................using the zero dimension...........they begin with eleven..............1 is always the numerator........................and any number on the bottom, the denominator..............starts with 1, for ever power of ten...........9 becomes 10, 99 becomes 100..........so anytime the denominator starts with a 1, which it does for any new power of ten, the number would start with an 11............................Gauss noticed a pattern in the primes.....for every power of 10............scaled down by 2.......there are gaps, skips.................jumps.............of different types.........

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:

${\displaystyle {\begin{aligned}&{}1+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}+{\frac {1}{5}}+{\frac {1}{6}}+{\frac {1}{7}}+{\frac {1}{8}}+{\frac {1}{9}}+\cdots \\[12pt]>{}&1+{\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{4}}+{\frac {1}{8}}+{\frac {1}{8}}+{\frac {1}{8}}+{\frac {1}{8}}+{\frac {1}{16}}+\cdots \end{aligned}}}$

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:

${\displaystyle {\begin{aligned}&{}1+\left({\frac {1}{2}}\right)+\left({\frac {1}{4}}+{\frac {1}{4}}\right)+\left({\frac {1}{8}}+{\frac {1}{8}}+{\frac {1}{8}}+{\frac {1}{8}}\right)+\left({\frac {1}{16}}+\cdots +{\frac {1}{16}}\right)+\cdots \\[12pt]={}&1+{\frac {1}{2}}+{\frac {1}{2}}+{\frac {1}{2}}+{\frac {1}{2}}+\cdots =\infty \end{aligned}}}$