Calculate $\sum_{ k = 1 } ^ n \ln ( 1 - x ^ k )$

Asked Jun 02 '20 at 20:33

Active Jun 03 '20 at 02:05

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I am trying to calculate the following sum $$f_n ( x ) = \sum_{ k = 1 } ^ n \ln ( 1 - x ^ k ), \qquad x \geq 0.$$ I first tried to use the power expansion of $\ln(1-u) = - \sum_{ m = 1 } ^ { \infty } \frac{ x^m } { m }$ but got stuck with infinite sums involving general terms of the form $$\frac{1}{m}\frac{x^m}{1 - x^m} \quad \mbox{and} \quad \frac{1}{m}\frac{x^{m(n+1)}}{1 - x^m}.$$ I also tried to compute the value of $f_n'(x)$, buth this time I got stuck with a finite sum involving a general term of the form $$\frac{k}{1 - x^k}.$$ If you have any idea, I would be happy to hear about it!

Thanks!

edited Jun 03 '20 at 02:05

an4s

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asked Jun 02 '20 at 20:33

Marie

Perhaps using $\ln A_1+ \ln A_2=\ln(A_1A_2)$ can be of some help. – Anurag A Jun 02 '20 at 20:40
1

Since $\ln(a) + \ln(b)\equiv\ln(a\cdot b)$, $\displaystyle\sum_{k = 1}^n\ln\left(1 - x^k\right)\equiv\ln\left(\prod_{k = 1}^n\left(1 - x^k\right)\right)$. Then, perhaps this is useful? This paper might be helpful. – an4s Jun 02 '20 at 20:53
that's related to q-Pochammer symbol – G Cab Jun 03 '20 at 00:00

Calculate $\sum_{ k = 1 } ^ n \ln ( 1 - x ^ k )$

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