Evaluating the sum after reordering an infinite series $1-\frac {1} {2}+\frac {1} {3}-\frac {1} {4}+\ldots $

I think you will find this result useful:

Fix $p$ and $q$, with $p\geq q\geq 1$. Then

$$\lim_{n \to \infty} \sum_{k=qn}^{pn} \frac{1}{k} = \log \frac{p}{q}$$

Usually, if you take the altered harmonic series and sum $p$ positive terms and then add $q$ negative terms, you'll get

$$\log 2 + \frac{1}{2} \log \frac{p}{q}$$

For example,

$$1+1/3+1/5-1/2-1/4+1/7+1/9+1/11-1/6-1/8+++--\cdots$$

will give

$$\log 2 + \frac{1}{2} \log \frac{3}{2}$$

So you now choose $p/q=a^2$

$$\log 2 + \frac{1}{2} \log a^2 =\log 2 + \log a = \log {2a}$$

Here is the germ of an idea: $$\frac12+\frac14+\frac14+\cdots+\frac1{2n}=\frac12\ln n+\frac\gamma2+o(1)$$ where $\gamma$ is Euler's constant, while the corresponding sum of odd reciprocals differs from the above by $\ln 2+o(1)$. I rather suspect that will do the trick.

By the way, I don't think the result is true if you allow reordering of the positive elements among themselves, or the negative ones. You should be able to reorder the harmonic series to diverge as slowly as you might wish by starting with $\sum 1/2^n$ and then inserting the reciprocals of non-powers of two at extremely rare intervals.