Proving pointwise convergence of series of functions Show that
$1/(1+x)+2x/(1+x^{2})+\cdots+mx^{m-1}/(1+x^m)+\cdots=1/(1-x)$
where $ m= 2^{n−1}$ and $−1 < x < 1$, in the sense of pointwise convergence.
I have tried to bound this by Wierestrass M-Test but haven't found one, also this test doesn't tell you about the limiting function. 
Another idea could be that if we can differentiate term by term on Right hand side and then check for the convergence of the differentiated series. If it converges then, the proof is done.
But differentiating the RHS is getting clumsier because $m$ is a function of $n$ and I am not getting any conclusive result. 
Please suggest me if there are easier ways to do this or even if it can be done the way I suggested, then please give an outline of proof. 
 A: Hint.  If you multiply out
$$(1+x)(1+x^2)(1+x^4)(1+x^8)\cdots\ ,$$
a "typical" term is
$$x^{a_1+a_2+\cdots+a_m}\ ,$$
where the $a_k$ are powers of $2$, with no power occurring more than once.  These expressions therefore give every $x^n$ (because every $n$ can be written as a binary expansion), and every $x^n$ only once (because the binary expansion is unique).  Therefore
$$(1+x)(1+x^2)(1+x^4)(1+x^8)\cdots=1+x+x^2+x^3+x^4+x^5+\cdots=\frac1{1-x}\ .$$
Now take logs and differentiate.
A: set $f_m(x) = 1+x^m$ so
$$
S(x) = \sum_{m=1}^{\infty} \frac{d}{dx} \log f_m(x) \\
= \frac{d}{dx} \sum_{m=1}^{\infty}\log f_m(x) \\
= \frac{d}{dx} \log \prod_{m=1}^{\infty} f_m(x) \\
= \frac{d}{dx} (-\log (1-x)) \\
= \frac1{1-x}
$$
A: Here is how you advance. Let's write the series in a compact form
$$ S(x) = \sum_{k=1}^{\infty}\frac{2^{k-1} x^{2^{k-1}-1}}{1+x^{2^{k-1}}} \implies \int_{0}^{x}S(t)dt = \sum_{k=1}^{\infty}\ln\left( 1+x^{2^{k-1}}\right) = \ln\prod_{k=1}^{\infty}\left( 1+x^{2^{k-1}} \right)$$
$$\implies \int_{0}^{x}S(t)dt = \ln(1-x) \implies = S(x)=\frac{1}{1-x}.  $$
