Infinite series equality $\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{3x^2}{1+x^3}+\frac{4x^3}{1+x^4}+\cdots$ Prove the following equality ($|x|<1$).
$$\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{3x^2}{1+x^3}+\frac{4x^3}{1+x^4}+\cdots\\
=\frac{1}{1-x}+\frac{3x^2}{1-x^3}+\frac{5x^4}{1-x^5}+\frac{7x^6}{1-x^7}+\cdots\\$$

Check that for $|x| < r < 1$, 
$$\frac{nx^{n-1}}{1+x^n}, \frac{nx^{n-1}}{1-x^n}, \frac{2nx^{2n-1}}{1-x^n}$$ are all $O(nr^n)$, which is summable.
Then, you can switch the order of summation in the following infinite sum : 
$$0 = \sum_{n \ge 1} (\frac{ nx^{n-1}}{1+x^n} - \frac {nx^{n-1}}{1-x^n} + \frac {2nx^{2n-1}}{1-x^{2n}})
= (\sum_{n \ge 1} \frac{ nx^{n-1}}{1+x^n}) - \sum_{n \ge 1}(\frac {nx^{n-1}}{1-x^n} - \frac {2nx^{2n-1}}{1-x^{2n}})$$
The second sum is a telescoping sum $\sum (a_n - a_{2n})$ so reordering again, it is $$(\sum a_n) - (\sum a_{2n}) = \sum_{n \text{ odd}} a_n + \sum_{n \text{ even}} (a_n - a_n) = \sum_{n\text{  odd}} a_n$$
Hence $$\sum_{n \ge 1} \frac{ nx^{n-1}}{1+x^n} - \sum_{n \text{ odd}}\frac {nx^{n-1}}{1-x^n}=0$$
