Proof of $\sum_{x = 1}^\infty \frac{1}{x}$'s divergence by absurdity?

Question

(From this site.)

The following argument purports to show that the series $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} \dots = 0$. It begins with the harmonic series.

$$ \begin{aligned} \sum \frac{1}{x} &= \sum \left( \frac{1}{2x - 1} + \frac{1}{2x} \right)\\ &= \sum \left( \frac{2(2x - 1)}{2x(2x - 1)} + \frac{1}{2x(2x - 1)} \right)\\ &= \sum \left(\frac{1}{x} + \frac{1}{2x(2x - 1)}\right)\\ &= \sum \left(\frac{1}{x}\right) + \sum\left(\frac{1}{2x(2x - 1)}\right)\\ \end{aligned} $$

So $\sum \frac{1}{x}$ equals itself plus something, which implies that the second series is $0$. Of course, this argument is false because the harmonic series diverges.

My question: suppose we didn't know the harmonic series diverged, but we did know that $\sum\left(\frac{1}{2x(2x - 1)}\right)$ converged to a nonzero value -- it is in fact $\ln{(2)}$.

Would we then be able to conclude, from the above argument, that the harmonic series diverges? That is, if we arrive at $A = A + c$, where $c$ is constant, can we conclude that $A$ is infinite?

Answer 1

If a series is absolutely convergent, then any rearrangement converges to the same value. Thus, assuming that $\sum \frac{1}{n}$ converges, we have absolute convergence (all the terms are positive).

Therefore all the operations made in your reasoning (grouping terms, splitting the series in two) are valid. Since you can write $\sum \frac{1}{n}$ as itself plus something positive, you get a contradiction.

Answer 2

$$A = A + c$$ so subtract A from both sides. $$c = 0$$ but $c$ is not $0$, it's $\ln 2$. Contradiction. $\therefore$ the harmonic series diverges.

Answer 3

$$ A = A + c \\ 1 = 1 + c/A \\ 0 = c/A \text{ only if A is infinite } $$