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 ($|x|<1$).
$$\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{3x^2}{1+x^3}+\frac{4x^3}{1+x^4}+\cdots\\
=\frac{1}{1-x}+\frac{3x^2}{1-x^3}+\frac{5x^4}{1-x^5}+\frac{7x^6}{1-x^7}+\cdots\\$$
 A: 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.
A: 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 $$\sum_{n \ge 1} \frac{ nx^{n-1}}{1+x^n} - \sum_{n \text{ odd}}\frac {nx^{n-1}}{1-x^n}=0$$
A: Multiplying $x$ to both sides, the identity is equivalent to
$$ \sum_{k=1}^{\infty} \frac{kx^k}{1+x^k} = \sum_{k=1}^{\infty} \frac{(2k-1)x^{2k-1}}{1-x^{2k-1}}. $$
Expanding and rearranging, each series can be written as
\begin{align*}
\sum_{k=1}^{\infty} \frac{kx^k}{1+x^k}
&= \sum_{k=1}^{\infty} \sum_{j=1}^{\infty} (-1)^{j-1} k x^{jk} = \sum_{n=1}^{\infty} \Bigg( \sum_{d|n} (-1)^{d-1}\frac{n}{d} \Bigg) x^{n} \\
\sum_{k=1}^{\infty} \frac{(2k-1)x^{2k-1}}{1-x^{2k-1}}
&= \sum_{k=1}^{\infty} \sum_{j=1}^{\infty} (2k-1)x^{j(2k-1)} = \sum_{n=1}^{\infty} \Bigg( \sum_{\substack{d | n \\ d \text{ odd}}} d \Bigg) x^{n}
\end{align*}
So it is sufficient to prove that
$$ \color{blue}{\sum_{d|n} (-1)^{d-1}\frac{n}{d} = \sum_{\substack{d | n \\ d \text{ odd}}} d} \quad \text{for } n = 1, 2, \cdots. \tag{1}$$
To this end, write $n = 2^e m$ for $e \geq 0$ and $m$ is odd. Then
$$ \sum_{d|n} (-1)^{d-1}\frac{n}{d}
= \sum_{d'|m} \underbrace{\sum_{i=0}^{e} (-1)^{2^i d'-1} 2^{e-i}}_{=1} \frac{m}{d'}
= \sum_{d'|m} \frac{m}{d'}
= \sum_{d|m} d
= \sum_{\substack{d | n \\ d \text{ odd}}} d $$
and hence $\text{(1)}$ is proved.
A: Let me add my own proof that I had in my mind.
$$Let\quad f(x)=\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{3x^2}{1+x^3}+\frac{4x^3}{1+x^4}+\frac{5x^4}{1+x^5}+\frac{6x^5}{1+x^6}+\frac{7x^6}{1+x^7}+\cdots$$
$$f(x)-\frac{1}{1-x}=\color{red}{\frac{-2x}{1-x^2}}+\frac{2x}{1+x^2}+\frac{3x^2}{1+x^3}+\frac{4x^3}{1+x^4}+\frac{5x^4}{1+x^5}+\frac{6x^5}{1+x^6}+\frac{7x^6}{1+x^7}+\cdots\\
=\frac{3x^2}{1+x^3}+\color{red}{\frac{-4x^3}{1-x^4}}+\frac{4x^3}{1+x^4}+\frac{5x^4}{1+x^5}+\frac{6x^5}{1+x^6}+\frac{7x^6}{1+x^7}+\cdots\\
=\cdots=\frac{3x^2}{1+x^3}+\frac{5x^4}{1+x^5}+\frac{6x^5}{1+x^6}+\frac{7x^6}{1+x^7}+\cdots+\lim_{n\to\infty}\frac{-2^nx^{2^n-1}}{1-x^{2^n}}
$$
$$f(x)-\frac{1}{1-x}-\frac{3x^2}{1-x^3}
=\frac{5x^4}{1+x^5}+\color{red}{\frac{-6x^5}{1-x^6}}+\frac{6x^5}{1+x^6}+\frac{7x^6}{1+x^7}+\cdots+\lim_{n\to\infty}\frac{-2^nx^{2^n-1}}{1-x^{2^n}}\\
=\cdots=\frac{5x^4}{1+x^5}+\frac{7x^6}{1+x^7}+\cdots+\lim_{n\to\infty}\frac{-2^nx^{2^n-1}}{1-x^{2^n}}+\lim_{n\to\infty}\frac{-3\cdot2^nx^{3\cdot2^n-1}}{1-x^{3\cdot2^n}}$$
$$\cdots$$
$$f(x)-\sum_{k=0}^{\infty}\frac{(2k+1)x^{2k}}{1-x^{2k+1}}=\sum_{k=0}^{\infty}\lim_{n\to\infty}\frac{-(2k+1)\cdot2^nx^{(2k+1)\cdot2^n-1}}{1-x^{(2k+1)\cdot2^n}}=0$$
$$\therefore f(x)=\sum_{k=0}^{\infty}\frac{(2k+1)x^{2k}}{1-x^{2k+1}}$$
