# Certain infinite series, arising from integral, equals zero

I want to prove that the following infinite series converges to 0: $$\sum_{i,j\ge 0} (-1)^{i+j}\frac {2i-j}{(j+1)(2i+1)(2i+j+2)} = \sum_{a,b \in N, a\equiv 1(2) } (-1)^{\frac{a-1}{2}+b-1}\frac{a-b}{ab(a+b)}$$

• Is it true? Yes. The problem arises from a certain integral that its value is 0.
• But what is the order of summation? Order matters since the series doesn't converge absolutely. The order is on increasing $2i+j$ (i.e., if there are $i,i',j, j'$ such that $2i+j<2i'+j'$, then $(i,j)$ comes before $(i',j')$.
• What is the integral it represents? $$\int_{0}^{1} \frac{\arctan x}{1+x} - \frac{\ln(1+x)}{1+x^2} dx = \int_{0}^{1} (f'g - fg') dx =$$ $$\sum_{i,j \ge 0, i+j \neq 0} \frac{i-j}{i+j} a_i b_j = \sum_{i,j\ge 0, i+j \neq 0} \frac{i}{i+j}(a_i b_j - b_i a_j)$$ where $$\ln(1+x) = f(x) = \sum a_i x^i, \arctan x = g(x) = \sum b_i x^i$$ The classical way of showing that it equals 0 is to first show that $\int_{0}^{1} \frac{\ln(1+x)}{1+x^2}dx = \frac{\pi}{8}\ln 2$ by change of variables ($t = \arctan x$) and certain symmetries, and then apply integration by parts. But I am interested in a new proof. Specifically, I want to find couples of terms in the series that have the same absolute value but opposite signs. It is not necessarily possible.

So my main motivation is this: I know that the claim is true, I started solving it in the stated way (different from classical integral tricks solutions), and I am stuck. Any help would be welcomed.

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I changed the \frac in your first to \sum as it was not rendering and it seems you want a sum. Please check that I got it right. –  Ross Millikan Jun 2 '11 at 18:11
@Ross: Thanks, I did made a type. I also removed an extra "1" in the numerator and made the indices nicer. –  Ofir Jun 2 '11 at 18:23
@Ross: When I looked at this page I saw the same error, so I instinctively edited it. Now, looking back at the history for some reason he undid your edit. @Ofir: Did you mean sum? I think it makes the most sense when the tex is that way. –  Eric Naslund Jun 2 '11 at 18:23
I am still new to the editing system and might have overwritten some of your changes accidentally since we were editing simultaneously. I meant summation, not a fraction. How do you check the editing history, by the way? –  Ofir Jun 2 '11 at 18:35