Can 2 be seen as 11??  The beg and end of the pf.......................a circle of sorts.......

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:

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

It follows (by the comparison test) that the sum of the harmonic series must be infinite as well. More precisely, the comparison above proves that

${\displaystyle \sum _{n=1}^{2^{k}}{\frac {1}{n}}\geq 1+{\frac {k}{2}}}$

for every positive integer k.
This proof, proposed by Nicole Oresme in around 1350, is considered by many in the mathematical community to be a high point of medieval mathematics. It is still a standard proof taught in mathematics classes today. Cauchy's condensation test is a generalization of this argument