I think it's fair to say that summing the series directly would be slow, even if it's not slow when you already happen to have summed the previous n-1 terms.

Not least because for modestly-sized target sums the number of terms you need to sum is more than is actually feasible. For instance, if you're interested in approximating a sum of 100 then you need something on the order of exp(100) or about 10^43 terms. You can't just say "well, it's not slow to add up 10^43 numbers, because it's quick if you've already done the first 10^43-1 of them".

It’s a natural observation, but it doesn’t address the floating point problem. I think the author should have said “fast or would accumulate floating point error” instead of “fast and would accumulate floating point error”.

You could compute in the reverse direction, starting from 1/n instead of starting from 1, this would produce a stable floating point sum but this method is slow.

Edit: Of course, for very large n, 1/n becomes unrepresentable in floating point.

There’s a trivial lower bound of zero, for n = 1.

For n > 1, there isn’t a lower bound. None of the numbers are integers again (https://en.wikipedia.org/wiki/Harmonic_series_(mathematics)#...), and because the difference between successive partial sums goes to zero and the series grows to arbitrary values, you’re bound to find a difference smaller than 1/(2n) somewhere beyond n.

No, because the terms tends monotonically towards zero. Let an integer m with closest harmonic number H_n be given (i.e. n minimizes |H_n-m|). So m exists either between H_n and H_(n+1) or H_n and H_(n-1). Then |H_n-m| < H_(n+1) - H_(n-1) = 1/n + 1/(n+1). We can make that bound arbitrary small by choosing a large enough n.

Spoiler: there is a simple argument against the existence of such a lower bound.