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 ($|x|<1$).
$$\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{3x^2}{1+x^3}+\frac{4x^3}{1+x^4}+\cdots\\
=\frac{1}{1-x}+\frac{3x^2}{1-x^3}+\frac{5x^4}{1-x^5}+\frac{7x^6}{1-x^7}+\cdots\\$$

Integrate both sides. Let L.H.S be $f(x)$ and R.H.S be $g(x)$.
(1). L.H.S
$$\int \frac{ix^{i-1}}{1+x^{i}} = \ln (1+x^i)$$
This gives $$\int f(x) dx= \sum_{i=1}^\infty \ln (1+x^i) = \ln (\prod_{i=1}^\infty (1+x^i))$$
(2). R.H.S
$$\int \frac{(2i+1)x^{2i}}{1-x^{2i+1}} = -\int \frac{-(2i+1)x^{2i}}{1-x^{2i+1}} = -\ln (1-x^{2i+1})$$
This gives $$\int g(x) dx=\sum_{i=0}^\infty -\ln (1-x^{2i+1}) = -\ln (\prod_{i=0}^\infty (1-x^{2i+1}))$$
(3). Finishing the problem.
Since $f(0)=g(0)=0$, it now suffices to show that $$ \prod_{i=1}^\infty (1+x^i) \cdot \prod_{i=0}^\infty (1-x^{2i+1}) = 1$$
Now here is the part that I'm not sure if it is true or not.
Let $$h(x)=\prod_{i=1}^\infty (1+x^i) \cdot \prod_{i=0}^\infty (1-x^{2i+1})$$
Of course, the domain of $h(x)$ is $|x|<1$.
Now we have $$h(x)=\prod_{i=1}^\infty (1+x^i) \cdot \prod_{i=0}^\infty (1-x^{2i+1})=\prod_{i=1}^\infty (1+x^{2i}) \cdot \prod_{i=0}^\infty (1+x^{2i+1}) \cdot \prod_{i=0}^\infty (1-x^{2i+1}) =\prod_{i=1}^\infty (1+x^{2i}) \cdot \prod_{i=0}^\infty (1-x^{4i+2})=h(x^2)$$
Therefore, for all $|x|<1$, we have $$h(x)=h(x^2)=h(x^4)= \cdots \lim_{n \to \infty} h(x^{2^n}) = h(0) = 1$$ as desired.
