I am reading some paper and I am slightly confused about the way the following product has been simplified by the author:

$$ (1-x)\ln(1-x) = (1-x)\left[-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-\cdots\right] $$

He then converts this into: $$ -x + \sum\limits_{l=1}^{\infty}\frac{x^{l+1}}{l(l+1)} $$

To obtain this, I tried multiplying the terms on R.H.S. in the first equation. However, I can apparently do it in two ways. First I can multiply by $1$ in the first bracket by the whole infinite expansion and then by second term ($-x$) to obtain: $$ -x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-\cdots +x^2+\frac{x^3}{2}+\frac{x^4}{3}+\frac{x^5}{4}+\cdots $$

Or I can multiply each term in the infinite series by $(1-x)$ to obtain:

$$ -x+x^2-\frac{x^2}{2}+\frac{x^3}{2}-\frac{x^3}{3}+\cdots $$ Which when simplified, gives the expression written by author. My question is, what makes the first way illegal and the second one legal? Or is it possible to show that both these methods are in fact the same thing? As a side note, first method seems illegal to me because the first series would never end.

  • 2
    $\begingroup$ Anything containing "$\ldots$" is illegal :) $\endgroup$ – Hagen von Eitzen Feb 2 '17 at 10:39
  • $\begingroup$ There is nothing illegal in the first notation. The first series is convergent so it has a limit, and so does the second. In addition, they are both absolutely convergent so that you can scramble them to the second way. When evaluating the series, just make sure that you compute the terms exponentially fast, so that it takes finite time in total :-) $\endgroup$ – Yves Daoust Feb 2 '17 at 11:41

$$(\color{red}1-x)\left(-x-\frac{x^2}2-\frac{x^3}3-\ldots\right)=-\color{red}x-\color{red}{\frac{x^2}2}-\color{red}{\frac{x^3}3}-\ldots+x^2+\frac{x^3}2+\ldots=$$ $$=-x+\frac{x^2}2+\frac{x^3}6+\frac{x^4}{12}+\ldots+\frac{x^{n+1}}{n(n+1)}+\ldots=-x+\sum_{n=1}^\infty\frac{x^{n+1}}{n(n+1)}$$



Any of the two proposed ways is legal when carried on within the convergence interval of the involved series, which is $\;|x|<1\;$ , as then we have absolute convergence and thus it is irrelevant the order in which we carry on the sum.

  • $\begingroup$ I am unsure about the second step. How can you combine the terms from red and black series? $\endgroup$ – Peaceful Feb 2 '17 at 10:42
  • $\begingroup$ @SnehalShekatkar One by one according to the powers of the $\;x\;$ , of course. Read the final part of the answer... $\endgroup$ – DonAntonio Feb 2 '17 at 10:44
  • $\begingroup$ Yes of course. However, the red series' terms would never end. So How can we at all combine the terms that come after "infinite number of terms"? Apparently, one of my maths professor has told me in the past that this cannot be done. $\endgroup$ – Peaceful Feb 2 '17 at 10:47
  • $\begingroup$ @SnehalShekatkar Do you know what "absolute convergence" is? $\endgroup$ – DonAntonio Feb 2 '17 at 10:49
  • $\begingroup$ Yes. If I understand correctly, if sum(|x_n|) converges then the series is absolutely convergent. $\endgroup$ – Peaceful Feb 2 '17 at 10:59

Do not fragment the terms, keep them as a sum. $$ (1-x)\ln(1-x) =(1-x)\left[-\sum_{n=1}^{\infty}\frac{x^n}{n}\right] =-\sum_{n=1}^{\infty}\frac{x^n}{n}+\sum_{n=1}^{\infty}\frac{x^{n+1}}{n} \\[8mm] =\left(-x-\sum_{\color{red}{n=2}}^{\infty}\frac{x^n}{n}\right)+\sum_{n=1}^{\infty}\frac{x^{n+1}}{n} =\left(-x-\sum_{\color{red}{n=1}}^{\infty}\frac{x^{n+1}}{n+1}\right)+\sum_{n=1}^{\infty}\frac{x^{n+1}}{n} \\[8mm] =-x+\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+1}\right)x^{n+1}=-x+\sum_{n=1}^{\infty}\frac{x^{n+1}}{n(n+1)} $$


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