Computing $ \sum\limits_{i=1}^{\infty}\sum\limits_{j=1}^{\infty} \frac{(-1)^{i+j}}{i+j}$ I would like to compute:
$$ \sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \frac{(-1)^{i+j}}{i+j}$$
I wanted to use Fubini's theorem for double series but $$ \frac{(-1)^{i+j}}{i+j}_{(i,j)\in\mathbb{N*^2}}$$ is not a summable family for $$ \forall i>0$$
$$ \sum_{j=1}^{\infty} \frac{1}{i+j}=\infty$$
What am I supposed to do?
 A: Let $S_n = \sum\limits_{i \geqslant 1, j \geqslant 1} \left( \frac{(-1)^{i+j}}{i+j} \mathbf{1}_{i+j \leqslant n+1} \right)$. Then
$$
   S_n = \sum_{i=1}^n \sum_{j=1}^{n+1-i} \frac{(-1)^{i+j}}{i+j} = 
     \sum_{i=1}^n \sum_{j=1}^{n+1-i}  \sum_{m=2}^{n+1} \delta_{i+j,m}\frac{(-1)^{i+j}}{i+j} =
     \sum_{m=2}^{n+1} \sum_{i=1}^n \sum_{j=1}^{n+1-i} \delta_{i+j,m}\frac{(-1)^{m}}{m}  =
     \sum_{m=2}^{n+1} (-1)^m\frac{m-1}{m} = \sum_{m=2}^{n+1} (-1)^m - \sum_{m=2}^{n+1} \frac{(-1)^m}{m} = \frac{1}{2}\left( 1 - (-1)^n \right) + \sum_{m=1}^n \frac{(-1)^m}{m+1} 
$$
Notice that
$$
   \lim_{n \to \infty} S_{2n} = \lim_{n \to \infty} \sum_{m=1}^{2n} \frac{(-1)^m}{m+1} = \lim_{n \to \infty} \left( \ln(2) - 1 + H_{n+1/2} - H_n \right) = \ln(2) - 1
$$
and
$$
  \lim_{n \to \infty} S_{2n+1} =  \lim_{n \to \infty} \left( 1 + \sum_{m=1}^{2n+1}  \frac{(-1)^m}{m+1} \right) = \lim_{n \to \infty} \left( 1 + \ln(2) - 1 +H_{n+1/2} - H_{n+1} \right) = \ln(2)
$$
Thus the sequence $S_n$ does not converges as $n$ increases, meaning that the original sum is not defined. The sequence $S_n$ has Cesaro mean, though:
$$
  \lim_{n \to \infty} \frac{1}{n} \sum_{m=1}^n S_m = \frac{1}{2} \left( \ln(2) + (\ln(2)-1)\right) = \ln(2) - \frac{1}{2}
$$
A: How about:
$$
-\int_0^1 (-x)^{i+j-1}\,dx = \frac{(-1)^{i+j}}{i+j}
$$
then for each $x \in (0,1)$ we have
$$
\sum_{i=1}^\infty\sum_{j=1}^\infty -(-x)^{i+j-1} = \frac{x}{(x+1)^2}
$$
and integrate
$$
\int_0^1\frac{x}{(x+1)^2}\,dx = \log 2 - \frac{1}{2} \approx 0.193147
$$
added
Explanation for summation inside integral ... Two uses of this nice "monotone alternating" convergence theorem:  Suppose $f_1(x) \ge 0\;$ is integrable on $E$ and $f_n(x) \searrow 0$ for almost every  $x \in E$. Then
$$
\sum_{n=1}^\infty (-1)^n \int_E f_n(x)\,dx = \int_E \left(\sum_{n=1}^\infty (-1)^n f_n(x)\right)\,dx
$$
PROOF: Group the terms in pairs.
added
More details now ...
$$
\int_0^1 -(-x)^{i+j-1}\,dx = \frac{(-1)^{i+j}}{i+j}
$$
For fixed $i$, the integrand decreases pointwise a.e. to zero in absolute value, and alternates sign.  Therefore
$$
\int_0^1 \sum_{j=1}^\infty-(-x)^{i+j-1}\,dx =
\sum_{j=1}^\infty\int_0^1 -(-x)^{i+j-1}\,dx =
 \sum_{j=1}^\infty \frac{(-1)^{i+j}}{i+j}
$$
Now this integrand is
$$
\sum_{j=1}^\infty-(-x)^{i+j-1} = \frac{-(-x)^i}{x+1}
$$
As $i$ varies, this decreases a.e. to zero in absolute value, and alternates sign, so
$$
\int_0^1 \sum_{i=1}^\infty \sum_{j=1}^\infty-(-x)^{i+j-1}\,dx=
\sum_{i=1}^\infty \int_0^1 \sum_{j=1}^\infty-(-x)^{i+j-1}\,dx=
\sum_{i=1}^\infty \sum_{j=1}^\infty \frac{(-1)^{i+j}}{i+j}
$$
