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

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