sum of the alternating harmonic series but every third term removed

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$$ X = \sum_{i=1}^{\infty} \frac{(-1)^{i+1}}{i} = \log 2$$

If we just look at the series formed from the third terms of the alternating harmonic series, we get

$$Y = -\frac{1}{3} + \frac{1}{6} - \frac{1}{9} + \frac{1}{12} +... $$ $$ = -\frac{1}{3} (1 - \frac{1}{2} + \frac{1}{3} - ...) = -\frac{\log 2}{3}$$

So one would think that the sum of the series in question is

$$ X - Y = \frac{4}{3}\log 2 = 0.9241962407465937$$

But numerical evaluation gives me one fourth of that $0.2310485601873915$.

What am I doing wrong?

Thanks