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

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

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

• The link in your post goes to page 327 with questions 463-507. Perhaps you ment to link to page 333? Nov 26, 2023 at 14:24
• 769 is prime :-) Nov 26, 2023 at 15:05
• @MartinSleziak, re, thanks for the corrected link! Since the question was better inlined anyway, I fixed the link while transcribing it. Nov 26, 2023 at 15:11
• If you like my answer, please accept it officially (so that it turns green). Thanks in advance! Nov 30, 2023 at 16:30

Observe that the left-hand side is the sum of two convergent series. Let $$N\geq 2$$ be an integer tending to infinity. Truncate the first series at the $$N$$-th term and the second series at the $$2N$$-th term. The resulting finite sum $$S_N$$ equals \begin{align*} S_N&=\sum_{2\leq n\leq N}\frac{\log 2}{n\log n\log 2n}+\sum_{2\leq n\leq 2N}\frac{(-1)^n}{n\log n}\\ &=\sum_{2\leq n\leq N}\left(\frac{1}{n\log n}-\frac{1}{n\log 2n}\right)+\sum_{\substack{2\leq n\leq 2N\\\text{n even}}}\frac{1}{n\log n} -\sum_{\substack{2\leq n\leq 2N\\\text{n odd}}}\frac{1}{n\log n}\\ &=\sum_{2\leq n\leq N}\frac{1}{n\log n}-2\sum_{\substack{4\leq n\leq 2N\\\text{n even}}}\frac{1}{n\log n}+\sum_{\substack{2\leq n\leq 2N\\\text{n even}}}\frac{1}{n\log n} -\sum_{\substack{2\leq n\leq 2N\\\text{n odd}}}\frac{1}{n\log n}\\ &=\sum_{2\leq n\leq N}\frac{1}{n\log n}+\frac{1}{2\log 2}-\sum_{\substack{4\leq n\leq 2N\\\text{n even}}}\frac{1}{n\log n}-\sum_{\substack{2\leq n\leq 2N\\\text{n odd}}}\frac{1}{n\log n}\\ &=\frac{1}{\log 2}+\sum_{3\leq n\leq N}\frac{1}{n\log n}-\sum_{\substack{4\leq n\leq 2N\\\text{n even}}}\frac{1}{n\log n} -\sum_{\substack{2\leq n\leq 2N\\\text{n odd}}}\frac{1}{n\log n}\\ &=\frac{1}{\log 2}+\sum_{3\leq n\leq N}\frac{1}{n\log n}-\sum_{3\leq n\leq 2N}\frac{1}{n\log n}\\ &=\frac{1}{\log 2}-\sum_{N Estimating the last sum crudely, it follows that $$\frac{1}{\log 2}-\frac{1}{\log N} In particular, $$\lim_{N\to\infty} S_N=\frac{1}{\log 2}.$$ Of course this limit equals the left-hand side in the original question, hence we are done.