In part (a) we are making a change of notation simply as a matter of convenience for the rest of this activity. The first few terms of the harmonic series are `1 + 1//2 + 1//3 + 1//4 + cdots `. The only reason these terms were numbered from `0` was because we obtained the series originally by substitution in a Taylor series. It is clearly simpler to start the numbering with `1`, in which case the sigma notation is `sum_(k=1)^(oo)1/k`.
For part (b), we observe that `1//3 > 1//4`, so
`1//3 + 1//4 > 1//4 + 1//4 =1//2`
and
`1 + 1//2 + 1//3 + 1//4 > 3//2 + 1//2 = 4//2.`
Continuing in this manner, we see that
`1//5 + 1//6 + 1//7 + 1//8 > 1//8 + 1//8 + 1//8 +1//8 = 1//2`,
so the sum of the first eight terms is more than `5//2`. Eight more terms add more than `1//2`, bringing the total for `16` terms above `6//2`. At each step we double the number of terms from the previous step, and we add more than `1//2`. To get the sum above `r//2`, we need `2^(r-2)` terms. No matter how large `r` is, we can always add up `2^(r-2)` terms, so there is no finite bound for the sum of the harmonic series. That is, the series diverges.
The harmonic series is an example of a series whose terms decrease to zero but whose sum nevertheless diverges to infinity. This does not contradict the Divergence Test because it is a one-way test. It says that a series whose terms do not approach zero must diverge, but it does not say anything about a series whose terms approach zero.