MHB How to Evaluate the Infinite Sum for 0 < x < 1?

[lfdahl]

For 0 < x< 1, find the sum:

\[\frac{2x-1}{1-x+x^2}+\frac{4x^3-2x}{1-x^2+x^4}+\frac{8x^7-4x^3}{1-x^4+x^8}+ ...\]

 

[Opalg]

[sp]\[\frac{2x-1}{1-x+x^2}+\frac{4x^3-2x}{1-x^2+x^4}+\frac{8x^7-4x^3}{1-x^4+x^8}+ \ldots\]
\[= \frac{\frac d{dx}(1-x+x^2)}{1-x+x^2} + \frac{\frac d{dx}(1-x^2+x^4)}{1-x^2+x^4}+\frac{\frac d{dx}(1-x^4+x^8)}{1-x^4+x^8}+ \ldots\]
\[= \frac d{dx}\bigl(\ln(1-x+x^2)\bigr) + \frac d{dx}\bigl(\ln(1-x^2+x^4)\bigr) + \frac d{dx}\bigl(\ln(1-x^4+x^8)\bigr) + \ldots\]
\[= \frac d{dx}\left(\ln\left(\frac{1+x^3}{1+x}\right) + \ln\left(\frac{1+x^6}{1+x^3}\right) + \ln\left(\frac{1+x^{12}}{1+x^6}\right) + \ldots\right)\]
\[= \frac d{dx}\left(\ln\left(\frac{1+x^3}{1+x}\frac{1+x^6}{1+x^3}\frac{1+x^{12}}{1+x^6} \cdots\right)\right)\]
\[= \lim_{n\to\infty} \frac d{dx}\left(\ln\left(\frac{1+x^{3\cdot2^n}}{1+x}\right)\right)\]
\[ = \frac d{dx}\left(\ln\left(\frac{1}{1+x}\right)\right)\]
\[= \frac d{dx}\bigl(-\ln(1+x)\bigr) = - \frac1{1+x}\]
Some of those steps may need heavy machinery to justify them (differentiating a series term by term, interchanging limits, ...). But that should all work satisfactorily for $0<x<1$.
[/sp]

 

[lfdahl]

[sp]\[\frac{2x-1}{1-x+x^2}+\frac{4x^3-2x}{1-x^2+x^4}+\frac{8x^7-4x^3}{1-x^4+x^8}+ \ldots\]
\[= \frac{\frac d{dx}(1-x+x^2)}{1-x+x^2} + \frac{\frac d{dx}(1-x^2+x^4)}{1-x^2+x^4}+\frac{\frac d{dx}(1-x^4+x^8)}{1-x^4+x^8}+ \ldots\]
\[= \frac d{dx}\bigl(\ln(1-x+x^2)\bigr) + \frac d{dx}\bigl(\ln(1-x^2+x^4)\bigr) + \frac d{dx}\bigl(\ln(1-x^4+x^8)\bigr) + \ldots\]
\[= \frac d{dx}\left(\ln\left(\frac{1+x^3}{1+x}\right) + \ln\left(\frac{1+x^6}{1+x^3}\right) + \ln\left(\frac{1+x^{12}}{1+x^6}\right) + \ldots\right)\]
\[= \frac d{dx}\left(\ln\left(\frac{1+x^3}{1+x}\frac{1+x^6}{1+x^3}\frac{1+x^{12}}{1+x^6} \cdots\right)\right)\]
\[= \lim_{n\to\infty} \frac d{dx}\left(\ln\left(\frac{1+x^{3\cdot2^n}}{1+x}\right)\right)\]
\[ = \frac d{dx}\left(\ln\left(\frac{1}{1+x}\right)\right)\]
\[= \frac d{dx}\bigl(-\ln(1+x)\bigr) = - \frac1{1+x}\]
Some of those steps may need heavy machinery to justify them (differentiating a series term by term, interchanging limits, ...). But that should all work satisfactorily for $0<x<1$.
[/sp]

Hi, Opalg

Spoiler

I´m afraid, there is an error in your calculus :(

There is not a telescoping product, because:

\[1-x+x^2 = \frac{1+x^3}{1+x} \\\\ 1-x^2+x^4 = \frac{1+x^6}{1+x^2} \\\\ 1-x^4+x^8 = \frac{1+x^{12}}{1+x^4} \\\\ 1-x^8+x^{16} = \frac{1+x^{24}}{1+x^8} \;\;...\]

 

[Olinguito]

[sp]Some of those steps may need heavy machinery to justify them (differentiating a series term by term, interchanging limits, ...). But that should all work satisfactorily for $0<x<1$.
[/sp]

Spoiler

IMHO the part that needs most justification is going from the fourth to the fifth line, where it’s assumed that
$$\sum_{n=1}^\infty\ln x_n\ =\ \ln\left(\prod_{n=1}^\infty x_n\right)$$
(which in turn assumes that the sum of the left and the product on the right converge).

 

[Opalg]

Hi, Opalg

Spoiler

I´m afraid, there is an error in your calculus :(

There is not a telescoping product, because:

\[1-x+x^2 = \frac{1+x^3}{1+x} \\\\ 1-x^2+x^4 = \frac{1+x^6}{1+x^2} \\\\ 1-x^4+x^8 = \frac{1+x^{12}}{1+x^4} \\\\ 1-x^8+x^{16} = \frac{1+x^{24}}{1+x^8} \;\;...\]

[sp]Oops, you're right, the product does not telescope at all! It should be
\[\frac d{dx}\left(\ln\left(\frac{1+x^3}{1+x}\frac{1+x^6}{1+x^2}\frac{1+x^{12}}{1+x^4}\frac{1+x^{24}}{1+x^8} \cdots\right)\right)\]
The denominators then form the product
\[(1+x)(1+x^2)(1+x^4)(1+x^8)\cdots = 1+x + x^2 + x^3 + x^4 + x^5 + x^6 + \ldots = \frac1{1-x}.\]
The numerators are the same, with $x^3$ instead of $x$, so their product is $\dfrac1{1-x^3}.$ Therefore we need to find
\[\frac d{dx}\left(\ln\left(\frac{1-x}{1-x^3}\right)\right) = \frac d{dx}\left(\ln\left(\frac1{1+x+x^2}\right)\right) = \frac d{dx}\bigl(-\ln(1+x+x^2)\bigr) = -\frac{2x+1}{1+x+x^2}.\]

I hope that works better than my first attempt.

[/sp]

 

[lfdahl]

[sp]Oops, you're right, the product does not telescope at all! It should be
\[\frac d{dx}\left(\ln\left(\frac{1+x^3}{1+x}\frac{1+x^6}{1+x^2}\frac{1+x^{12}}{1+x^4}\frac{1+x^{24}}{1+x^8} \cdots\right)\right)\]
The denominators then form the product
\[(1+x)(1+x^2)(1+x^4)(1+x^8)\cdots = 1+x + x^2 + x^3 + x^4 + x^5 + x^6 + \ldots = \frac1{1-x}.\]
The numerators are the same, with $x^3$ instead of $x$, so their product is $\dfrac1{1-x^3}.$ Therefore we need to find
\[\frac d{dx}\left(\ln\left(\frac{1-x}{1-x^3}\right)\right) = \frac d{dx}\left(\ln\left(\frac1{1+x+x^2}\right)\right) = \frac d{dx}\bigl(-\ln(1+x+x^2)\bigr) = -\frac{2x+1}{1+x+x^2}.\]

I hope that works better than my first attempt.

[/sp]

Thankyou for participating in this challenge, Opalg! Yes sure, your 2nd attempt works perfect!

 

[lfdahl]

Spoiler

IMHO the part that needs most justification is going from the fourth to the fifth line, where it’s assumed that
$$\sum_{n=1}^\infty\ln x_n\ =\ \ln\left(\prod_{n=1}^\infty x_n\right)$$
(which in turn assumes that the sum of the left and the product on the right converge).

Hi, Olinguito

Spoiler

Thankyou for your sharp observation in Opalg´s solution concerning convergence.
Do you perhaps have any idea of how to prove convergence for both sides of the equation? Or is it sufficient to show, that we´re dealing with two geometric series, and that $0<x<1$?