Sign of coefficients

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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I think $a_2=-\frac{1}{12}$, $a_3=-\frac{1}{24}$, $a_4=-\frac{19}{720}$. – GH from MO Apr 29 '13 at 11:32
@GH: yes, sorry, untidy notes... – Anton Apr 29 '13 at 11:57

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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See also oeis.org/A006232 -- if the OP had had tidier notes, he might have found this on his own. ;-) – Barry Cipra Apr 29 '13 at 14:20

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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I knew this equation, I started with it. Computing of iterated differentials didn't help. – Anton Apr 29 '13 at 11:56
Ok, sorry about my sloppy answer ;) – Loïc Teyssier Apr 29 '13 at 15:34