Does the harmonic series converge or diverge?

I have been doing some work (well a LOT of work) on brushing up a variety of precalculus and calculus topics (i.e. every one I can find on the website) on Khan Academy recently. This involved a huge amount of stuff I have done before, but usually with additional insights (such as logarithmic implicit differentiation) and new techniques (e.g. L'Hôpital's Rule for the two-sided limit of a function, and using the natural logarithm to turn exponential problems into rational functions for said rule to be applicable). But in my fascinatingly different learning on divergence/convergence tests for infinite series, something prickly came up.

As Sal Khan informs the viewer, quite correctly of course, there is a famous proof from French Medieval mathematician Nicole Oresme showing that the infinite harmonic series is divergent. It does this by way of the comparison divergence/convergence test. So first off, what is this test?

The direct comparison test involves the infinite series under test, S_A, and a second infinite series S_B. Every term of the series S_A is smaller than the corresponding term in  S_B, and all terms are non-negative. The statements are thus: if S_B converges, then S_A must also converge; if S_A diverges, then S_B must also diverge.

Both of these statements are intuitively easy to grasp. Upon reading this, I immediately questioned the use of 'must converge' versus 'is bounded', simply because the counterexample that sprung to mind was a sequence containing the sine or cosine functions, or any other type of oscillatory function such as one including the term (-1)ⁿ - here it would be ok to say it is bounded if the first condition was satisfied, yet the very nature of the function would potentially stop a true limit being ever reached. However, in reality although this is true it would be a trivial case to use, since it defeats the premise of actually using the direct comparison test if you already know the function is divergent. And this is forgetting the possible exception that the oscillator is appended to a decaying function, in which case the problem would sort itself by zeroing out as n approaches infinity. Nevertheless the test would still be useful for determining whether said function is bounded or not, in fairness.

So, moving onto the proof. Oresme took the series S_A = 1 + ¹/₂ + ¹/₃ + ¹/₄ + ... (infinite harmonic series) and for each term found the smallest power of  ¹/₂ that was smaller than or equal to it. This led to this comparison:

S_A = 1 + ¹/₂ + ¹/₃ + ¹/₄ + ¹/₅ + ¹/₆ + ¹/₇ + ¹/₈ + ¹/₉ + ...
S_B = 1 + ¹/₂ + ¹/₄ + ¹/₄ + ¹/₈ + ¹/₈ + ¹/₈ + ¹/₈ + ¹/₁₆ + ...

Since each term in S_B is smaller than the corresponding term in S_A, it must be that the value of S_A up to the nth term is smaller than the value of S_B at the same point. So we need to determine the behaviour of S_B in the limit as n stretches off to infinity, in order to perform the direct comparison test.

It is easy to notice that the terms in S_B can be grouped in a very special way - there are two quarters, four eighths, eight sixteenths etc. This leads to the clearly divergent series:

S_B = 1 + ¹/₂ + ¹/₂ + ¹/₂ + ...

Since this is divergent, it must also be that the harmonic series is divergent too. This is nicely illustrated on Desmos, with the caveat that functions based on discrete summations of integral-domained terms plot discrete line segments centered about each integer. So it looks like a bunch of lines when in reality is is a set of discrete points. Anyway the visual representation is sound:

It's funny that out of all of that, the thing I'm most proud of it that I have finally found a vaguely practical use for the floor function! What it is doing in the red line (representing S_B in the diagram) is finding the exact value of the power 0.5 needs to be raised to to equal the corresponding term in the blue line, then rounding it up to the next integer value.

Ok so that looks pretty convincing - the red line diverges so it stops the blue line converging too. It just seems very counterintuitive though. Although failing the nth term divergence test doesn't tell you for sure that a series converges, it would seem highly logical that the harmonic series should converge since its terms will get infinitely small in the limit! So for me, I want even more argument to convince myself that this is true.

The next thing we can do is to analyse how quickly the comparison function, S_B, is growing. Obviously the rate of change decreases as n increases, because the additional increments drop in powers of 2 at a time. But the number of terms between each drop also increases by a factor of 2 each time (consider there are 2 terms of ¹/₄, 4 terms of ¹/₈, 8 terms of ¹/₁₆...). So, for example, when the gradient of the line is ¹/₁₂₈ (where 128 = 2⁷), 2⁷ < n < 2⁸, or more clearly 128 < n < 256. More generally, we find that for some natural number k, the region in which the gradient is ¹/_2^k is 2^k < n < 2^k+1. All this still doesn't help me reason this out though, because it would still suggest that as x → ∞, the gradient of the line → ¹/_∞ = 0.

In conclusion, it is very difficult to condense the reasoning to anything more satisfying than a series comparison at my current level of knowledge. Keep studying then...