How to prove the following?
$$ \sum _{k \geq 1} \log \left(\frac{1}{1-z^k}\right)=\sum _{k,t\geq1} \frac{z^{k t}}{t} $$
Source: Knuth's TAOCP (Vol.1, answer to ex. 2.3.4.4, 1.), where it's mentioned casually without explanation.
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As $\ln(1-x)=-\sum_{r=1}^\infty\dfrac{x^r}r$ for $-1\le x<1$
$\ln\dfrac1{1-z^k}=-\ln(1-z^k)=\sum_{t=1}^\infty\dfrac{(z^k)^t}t$ assuming $-1\le z^k<1$
lab bhattacharjee
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See http://math.stackexchange.com/questions/594838/taylor-series-for-log1x-and-its-convergence http://math.stackexchange.com/questions/322600/uniform-convergence-of-power-series-of-log1x-on-0-1 http://math.stackexchange.com/questions/365516/why-does-the-taylor-series-of-ln-1-x-only-approximate-it-for-1x-le-1 – lab bhattacharjee Jan 14 '16 at 16:43
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For $|x| < 1$, we have $$\frac {1} {1 - x} = \sum_{t \geqslant 0} x^t.$$ Integrating both sides in respect to $x$, we get $$\log \frac {1} {1 - x} = \sum_{t \geqslant 1} \frac {x^t} {t}.$$ If we set $x = z^k$ for $k = 1, 2, \cdots$ and then sum, we will get $$\sum_{k \geqslant 1} \log \frac {1} {1 - z^k} = \sum_{k, t \geqslant 1} \frac {z^{kt}} {t},$$ as desired.