Gdddk

The proof involving partial sums up to the nth term, where n is some power of $2$, completely makes sense. But just looking at the series itself, it seems very strange that it's divergent.

For large values of $n$, $a_n$ would start being extremely small and having an indistinguishable effect on the overall sum. All the sixth sense I've gained from working with limits makes it seem really strange that this would be considered divergent.

Surely there is a number (not even that difficult to find) such that we don't have enough computational power to calculate it's difference with the next terms (seeing as we'd be calculating differences based on hundreds of decimal places).

If you've any intuition on this I'd very much love to hear it!

Edit: I'm not asking for the proof of why it's divergent, I'm asking for peoples' personal ways of thinking and making sense of this intuitively. The post suggested to have been duplicated presents formal proofs; that's not what I'm looking for :)

From Real Infinite Series by Bonar:

We know that, for $x>-1$, $$x geq ln(1+x)$$ Now $$sum_1^n frac1kgeq sum_1^n lnleft(1+frac1kright)=ln(n+1) longrightarrow infty$$ as $n to infty$ and hence the divergence of the harmonic series follows

We can interpret this argument in a much more strikingly visual way as follows:

Consider the following graph of the function $g(x) = sin(pi e^x)$, shown below, We consider $g$ as a function only of positive reals, We know that this function is defined for arbitrarily large $x$. We also know that $sin x$ is zero at integer multiples of $pi$, so that $g$ has zeros whenever $e^x$ is integer-valued, which happens of course for $x$ of the form $log n$. The distance between consecutive zeros is of the form $log(k + 1) — log k$, which by the argument above is a lower bound to $1/k$. This is the motivation for the choice of the function $g$—the oscillations make visible the segments between zeros, and the lengths of these segments estimate the terms of the harmonic series. If the harmonic series were to converge to some number $N$, then the length sum of all the segments between zeros of $g$, since they are smaller, would also be bounded above by $N$. Then $g$ could have no further zeros right of the vertical line $x = N$, but we know this does not happen. Again we emphasize that this contains no mathematical content not present in the argument above, only a new way to make it tangible.

Added: Also the author of the above mentioned book gives $11$ proofs of " $sum frac1n$ is divergent". So refer this book for more details!

Variation of the proof cited by the OP: let be $S_n$ the $n$-th partial sum:
$$S_2n - S_n = sum_k=n+1^2nfrac1kge(2n - n)frac12n = frac12.$$
(lower bound: number of terms times smallest term)