I am not sure.........................but u can def. find the sum of the gaps........between jumps.........from any time the denominator jumps to the next power of ten.......like the 1st jump......1/9 + 1/10...........all numbers below 1/10........say or until the next jump.....1/100..............it would go......1/99 + 1/100..............so u CAN compute.............1/10 to 1/100..........................like Gauss did......it might not be exact........but u can get a rough estimate.........

Divergence[edit]

There are several well-known proofs of the divergence of the harmonic series. A few of them are given below.

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:

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