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How could one prove that

$$x = \sum_{k=2}^\infty k \left( \sum_{j=2^k}^{2^{k+1}-1} \frac{(-1)^j}{j} \right )$$ is such that $x+{1\over2}-{1\over3} = \gamma$ ?

I am having problems just calculating to see if it is correct, let alone proving it...

Amr
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JohnWO
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  • In that case, where did you get this? – Will Jagy Jun 18 '13 at 01:45
  • @WillJagy: To what case does "that" refer to? – JohnWO Jun 18 '13 at 01:52
  • You do not know whether this is correct, nor can you calculate it. Yet you have a suspicion that it is correct. Someone or something told you it was true, or at least likely and worth investigating. You have not identified the someone or something. – Will Jagy Jun 18 '13 at 02:20
  • @WillJagy: It is a conjecture, which I want to prove, based on an interpretation of the Ramanujan expansion. I can show that it's close, by the means of calculating it, but I have no proof for it. – JohnWO Jun 18 '13 at 02:37
  • Why don't you write the sum on the left and $\gamma -\frac{1}{6}$ on the right? – John Bentin Jun 18 '13 at 06:34
  • @JohnBentin: You say təˈmeɪtoʊz, I say təˈmɑːtoʊz... Let's Call the Whole Thing Off... – JohnWO Jun 18 '13 at 10:47
  • I would love to see this proven in a way other than proof by exhaustion. – JohnWO Jun 18 '13 at 10:51
  • @JohnBentin: the $\frac12-\frac13$ is just the $k=1$ term. My question is why not simply extend the sum to $k=1$ and eliminate the $\frac12-\frac13$? – robjohn Jun 18 '13 at 11:12
  • @robjohn: That would help in answering the question; Please share. – JohnWO Jun 18 '13 at 11:28
  • @JohnWO: I don't understand what you want me to share. I've answered the question. I was simply wondering why the $k=1$ term was brought out of the outer sum as $\frac12-\frac13$. – robjohn Jun 18 '13 at 12:26
  • @robjohn: Yes, that's simplest. – John Bentin Jun 18 '13 at 13:39

1 Answers1

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Start with equation $(15)$ in this answer: $$ \begin{align} \gamma &=\sum_{k=1}^\infty AHT(2^k-1)\\ &=\sum_{k=1}^\infty\sum_{j=0}^\infty\frac{(-1)^j}{2^k+j}\\ &=\color{#C00000}{\sum_{k=1}^\infty\sum_{j=2^k}^\infty\frac{(-1)^j}j}\\ &=\sum_{k=1}^\infty\sum_{i=k}^\infty\sum_{j=2^i}^{2^{i+1}-1}\frac{(-1)^j}j\\ &=\sum_{i=1}^\infty\sum_{k=1}^i\sum_{j=2^i}^{2^{i+1}-1}\frac{(-1)^j}j\\ &=\color{#C00000}{\sum_{i=1}^\infty i\sum_{j=2^i}^{2^{i+1}-1}\frac{(-1)^j}j}\\ &=\frac12-\frac13+\sum_{i=2}^\infty i\sum_{j=2^i}^{2^{i+1}-1}\frac{(-1)^j}j \end{align} $$

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