Walking on logs

The distance the man travels in the positive $x$ -direction is

$\displaystyle\sum_{k=1}^{\infty} (-1)^{k+1}\ln\left(\dfrac{2k + 6}{2k + 2}\right) =$

$\displaystyle\sum_{k=1}^{\infty} (-1)^{k+1} \ln(2(k + 3)) - \sum_{k=1}^{\infty} (-1)^{k+1} \ln(2(k + 1)) =$

$\displaystyle\sum_{k=3}^{\infty} (-1)^{k+1} \ln(2(k + 1)) - \sum_{k=1}^{\infty} (-1)^{k+1} \ln(2(k + 1)) =$

$-\ln(4) + \ln(6) = \ln\left(\dfrac{3}{2}\right),$

where the last bit of index shifting was done to reveal that all the terms canceled pairwise except the first two terms of the second sum. The same indexing "trickery" can be employed in determining the distance traveled in the positive $y$ -direction, which will be $-\ln(5) + \ln(7) = \ln\left(\dfrac{7}{5}\right).$

The magnitude of the displacement of the man from the origin will then be

$\sqrt{\left(\ln\left(\dfrac{3}{2}\right)\right)^{2} + \left(\ln\left(\dfrac{7}{5}\right)\right)^{2}} = 0.5269$ to 4 decimal places.

To 2 decimal places, the desired value is then $\boxed{0.53}.$