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The function $\frac{x}{\ln(1-x)}$ has a Taylor series $-1+c_1x+c_2x^2+\cdots$ and I want to show $c_1>c_2>\cdots>0$. More generally, is there a result about how a function has positive Taylor coefficients or has decreasing Taylor coefficients?

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  • $\begingroup$ I already found the recurrence relation $c_n=\frac{1}{n+1}-\frac{1}{n}c_1-\cdots -\frac{1}{2}c_{n-1}$ but it seems to be nothing. $\endgroup$
    – Bo WANG
    Commented Aug 11, 2015 at 3:04
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    $\begingroup$ The numbers appear to be the same as those tabulated at oeis.org/A002206 where there are many links and references (this does not address the more general question). $\endgroup$ Commented Aug 11, 2015 at 3:06

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We have $$-\int_{-1}^0 {1\over {(z+1)^t }}dt={ {-z}\over{\log(1+z)}}, \ \ \ \int_{-2}^{-1}{1\over {(z+1)^t }}dt={{z(z+1)}\over{\log(1+z)}}, \ \ z>0$$ So the k-derivative of these functions is positive when k is even and negative when k is odd. So the coefficients of the Taylor series of ${z\over{\log(1-z)}},{{z^2-z}\over{\log(1-z)}}$ are positive, as desired.

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  • $\begingroup$ Wow! Pretty elegant proof! $\endgroup$
    – Bo WANG
    Commented Aug 16, 2015 at 15:55
  • $\begingroup$ How does this show the coefficients are decreasing? $\endgroup$
    – japalmer
    Commented Jul 2 at 16:12

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