Harmonic series

Take portions of successive powers of two, beginning with 1/2. That is, take $[\frac{1}{2}],[\frac{1}{3},\frac{1}{4}],[\frac{1}{5},\frac{1}{6},\frac{1}{7},\frac{1}{8}]...$ , such that a portion of size $2^i$ consists of $[\frac{1}{2^i+1} ... \frac{1}{2^{i+1}}]$ .

It can be seen that the sum of each portion is greater than $\frac{1}{2}$ : $\frac{1}{2^i+1}+ ... \frac{1}{2^{i+1}}> \frac{1}{2^{i+1}} + ... \frac{1}{2^{i+1}}$ $= \frac{2^i}{2^{i+1}}$ $=\frac{1}{2}$

Indeed, if there exists an upper-bound $M$ , then by the $2\lceil M \rceil$ portion, the harmonic series will be greater than $\frac{1}{2}*2\lceil M \rceil =\lceil M \rceil \geq M$ , contradicting the definition of the bound.

There are infinite number of primes. Thus in the above sum there are infinite prime denominators. Thus you can construct infinite many geometric series by rearranging with rates 1/p. All these series converge to some positive value. Since there are infinite number of values, the series must diverge.

One can prove it by showing that the given sequence of partial sums is not a Cauchy sequence, hence not convergent

Σ(n=1,infinity)(1/n)=lim(k->infinity)(Σ(n=1,k)(1/n))=lim(k->infinity)((1/k)Σ(n=1,k)(1/(n/k)))=integral(x=0,1)(1/x×dx)=(ln|x|)(x=0,1)=+infinity => diverges