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Let $a_0,a_1,\dots$ be the sequence satisfying $$ \left(\sum_{n=0}^\infty a_n x^n\right)\left(\sum_{n=0}^\infty \frac{x^n}{n+1}\right)=1. $$ This means that $a_0=1$ and $a_{n+1}=-\sum_{j=0}^n\frac{a_j}{n+2-j}$. One gets $a_1=-\frac12$ and $a_3=-\frac{13}{720}$.

The first and most important question is: Is it true that all coefficients $a_n$ for $n\ge 1$ are strictly negative?

Secondly, is there any connection to, say, Bernoulli numbers or any other sequence that carries a name?

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  • $\begingroup$ I think $a_2=-\frac{1}{12}$, $a_3=-\frac{1}{24}$, $a_4=-\frac{19}{720}$. $\endgroup$
    – GH from MO
    Apr 29, 2013 at 11:32
  • $\begingroup$ @GH: yes, sorry, untidy notes... $\endgroup$
    – user1688
    Apr 29, 2013 at 11:57

2 Answers 2

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These numbers (with alternating signs) are called Bernoulli numbers of the second kind or Cauchy numbers. Two proofs that these numbers are negative can be found in http://people.brandeis.edu/~gessel/homepage/slides/analysis-nec.pdf.

Another reference is

Merlini, Donatella; Sprugnoli, Renzo; Verri, M. Cecilia, The Cauchy numbers. Discrete Math. 306 (2006), no. 16, 1906–1920.

The determination of the sign of these numbers can be found much earlier in Charles Jordan's Calculus of Finite Differences, page 267.

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    $\begingroup$ See also oeis.org/A006232 -- if the OP had had tidier notes, he might have found this on his own. ;-) $\endgroup$ Apr 29, 2013 at 14:20
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Since the second series can be explicitely computed, you have $$\sum_{n=0}^{\infty}a_n x^n = -\frac{x}{\ln(1-x)}.$$ Now you can compute explicitely a recurrence relation giving you the exact value of $a_n$ by computing iterated differentials at $0$ of this analytic function. Or derive explicitely the sign of $a_n$, which I believe to be not too difficult, but I don't have the time right now to do it…

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  • $\begingroup$ I knew this equation, I started with it. Computing of iterated differentials didn't help. $\endgroup$
    – user1688
    Apr 29, 2013 at 11:56
  • $\begingroup$ Ok, sorry about my sloppy answer ;) $\endgroup$ Apr 29, 2013 at 15:34

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