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.

• What "$= 2n-1$"? – marty cohen Feb 19 '15 at 2:18
• Corrected, it should be $m=2^{n-1}$ – Silver moon Feb 19 '15 at 2:20
• So is your series then.. $$\sum_{n=1}^{\infty} \dfrac{2^{n-1} x^{2^{n-1}-1}}{1+x^{2^{n-1}}}$$ – Cameron Williams Feb 19 '15 at 2:22
• Sorry, accidentally hit enter. See my edited comment. – Cameron Williams Feb 19 '15 at 2:24
• Is the third term $3x^2/(1+x^3)$ or $(4x^3)/(1+x^4)$? – marty cohen Feb 19 '15 at 2:26

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.

• Are the conditions satisfied which allow differentiation? – Silver moon Feb 19 '15 at 2:41
• Not sure, but since you are only asked for pointwise convergence you can regard $x$ as fixed and work with the finite version of the product,$$(1+x)(1+x^2)(1+x^4)\cdots(1+x^{2^{m-1}})=\frac{1-x^{2^m}}{1-x}\ .$$Give it a try. – David Feb 19 '15 at 2:58

set $f_m(x) = 1+x^m$ so

$$S(x) = \sum_{m=1}^{\infty} \frac{d}{dx} \log f_m(x) \\ = \frac{d}{dx} \sum_{m=1}^{\infty}\log f_m(x) \\ = \frac{d}{dx} \log \prod_{m=1}^{\infty} f_m(x) \\ = \frac{d}{dx} (-\log (1-x)) \\ = \frac1{1-x}$$

Here is how you advance. Let's write the series in a compact form

$$S(x) = \sum_{k=1}^{\infty}\frac{2^{k-1} x^{2^{k-1}-1}}{1+x^{2^{k-1}}} \implies \int_{0}^{x}S(t)dt = \sum_{k=1}^{\infty}\ln\left( 1+x^{2^{k-1}}\right) = \ln\prod_{k=1}^{\infty}\left( 1+x^{2^{k-1}} \right)$$

$$\implies \int_{0}^{x}S(t)dt = \ln(1-x) \implies = S(x)=\frac{1}{1-x}.$$

• How can you interchange limits with respect to integration? – Silver moon Feb 19 '15 at 3:04
• Doing this requires uniform convergence of the series, doesn't it? – Silver moon Feb 19 '15 at 3:04
• @Martin: You should justify some of the steps that I did. Like interchanging the sum with the integral for instance. I assume you know about uniform convergence and related issues! – science Feb 19 '15 at 3:08
• Yeah I know. I have mentioned about uniform convergence in my comment as well for interchanging limits. I just wanted to ask if you have checked for these issues. I will surely justify it for me. – Silver moon Feb 19 '15 at 3:53
• I have an objection to your first step. In the first step you assume that the given series converges uniformly to $S(x)$ for each $x\in [a,b]$ where $a>-1$ and $b<1$ and then you pass on to the limits and integrate termwise (Otherwise, this operation is not valid). But how do you know that the series has a uniform convergence to $S(x)$? Apart from that everything looks okay. Please justify. – Silver moon Feb 19 '15 at 7:57