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How to show that $\log 2(1/2\log 2\log 4 + 1/3\log 3\log 6 + \dotsb) + 1/2\log 2 - 1/3\log 3 + 1/4\log 4 - \dotsb = 1/\log 2$

I've been studying Ramanujan's work and I stumbled upon this question in the book: Collected Papers of Srinivasa Ramanujan. In there I found question number 769 which is about an infinite sum with logarithms. I didn't found a solution, so can someone please explain it to me:

Show that

$$\log 2\left(\dfrac1{2\log 2\log 4} + \dfrac1{3\log 3\log 6} + \dfrac1{4\log 4\log 8} + \dotsb\right) + \dfrac1{2\log 2} - \dfrac1{3\log 3} + \dfrac1{4\log 4} - \dfrac1{5\log 5} + \dotsb = \dfrac1{\log 2}.$$