# Question about proof the harmonic series diverges

The usual proof that the harmonic series diverges that I have seen involves group together terms that sum to $$\frac{1}{2}$$. It may take the form: $$\sum\limits_{n=1}^{\infty} \frac{1}{n} \ge 1 + \frac{1}{2} + \left(\frac{1}{4} + \frac{1}{4}\right) + \left(\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}\right) + \ldots$$ Here is my question: why can I even write that down (or can I write that down?) when we don't have an associative law for an infinite sum? If the series converges absolutely, I should be allowed to rearrange and regroup terms, but for a divergent series, why can I invoke associativity?

(I suppose the proof is technically possible even if I didn't group terms just by acknowledging that if I sum them in this way, I get $$\frac{1}{2}$$ each time. If that is the point, and the "grouping" is nothing more than a way for us to see the pattern, I am fine with this.)

That's a pretty good question! The way I would resolve it is to remember that, by definition, an infinite sum is the limit of the sequence of partial sums: $$\sum_{n=1}^\infty \frac1n = \lim_{N\to\infty} \bigg( \sum_{n=1}^N \frac1n \bigg).$$ What this proof is doing is having us notice that for any $$k\ge0$$, $$\sum_{n=1}^{2^k} \frac1n \ge 1 + \sum_{j=1}^k \bigg( 2^{j-1} \text{ copies of } \frac1{2^j} \bigg) = 1+\frac k2.$$ Therefore $$\limsup_{n\to\infty} \bigg( \sum_{n=1}^N \frac1n \bigg) \ge \limsup_{k\to\infty} \bigg( \sum_{n=1}^{2^k} \frac1n \bigg) = \limsup_{k\to\infty} \bigg( 1+\frac k2 \bigg) = \infty,$$ which is enough to show that the series diverges. (Indeed, since all of the summands are nonnegative, it's easy to show that the limit exists and equals the lim sup.)
In one sense, this sort of proof does show that associativity holds for infinite series, as long as that series converges or diverges to $$\pm\infty$$. But since (as you point out) one usually doesn't know that in advance, maybe it's better to be concrete in any particular case at first.