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I have a particular sequence, the coefficients determined by the generating function: $$\frac{2e^x}{e^{2x}+1+2x}=\sum_{n=0}^\infty\varepsilon_n\frac{x^n}{n!}$$ The first few numbers are $$1,-1,3,-15,93,-725,6815,-74627,...$$ I was looking for patterns amongst the digits so I naturally looked at the ratios of consecutive digits. Using mathematica I got the sequence: $$-1., -3., -5., -6.2, -7.7957, -9.4, -10.9504, -12.5136, -14.0798, -15.6438, -17.2081, $$ $$-18.7725, -20.3369, -21.9013, -23.4656, -25.03, -26.5944, -28.1588, $$ $$-29.7232, -31.2875, -32.8519, -34.4163, -35.9807, -37.545, -39.1094, -40.6738,$$ $$ -42.2382, -43.8025, -45.3669, -46.9313, -48.4957, -50.0601, -51.6244, -53.1888, $$ $$-54.7532, -56.3176, -57.8819, -59.4463, -61.0107, -62.5751, -64.1394, -65.7038, $$ It appeared that while the ratios weren't converging, if I took their difference, there was a convergence of approximately -1.5. So I took their difference in mathematica out to about 300 terms and I got a number, seemingly irrational, (since the further I went into the sequence the decimal expansion got longer and longer). I was wondering how to show that it is irrational, but I was really looking to see if the number is recognizable as any particular known number. So, I guess I'm looking more for a reference request as to where, if any, I could find out whether or not the number is a known irrational value, and if it is, how it arises. The number, taken to 50 digits here is

$$ -1.5643765885603998024405941518534895603638168079326... $$ The numbers alternate so I always get a negative.

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    $\begingroup$ Due to the form of the denominator, it appears unlikely that the limit lies in $\mathbb Q[\pi,e]$. Perhaps something can be done using special values of the Lambert W function. $\endgroup$ – pre-kidney Jun 11 '15 at 1:56
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Heuristic. Assume that the Taylor expansion of the function has the form

$$ f(z) = \frac{2e^{z}}{e^{2z} + 1 + 2z} = \sum_{n=0}^{\infty} (-1)^n a_n z^n, $$

where $a_n$ are positive as predicted by OP's calculation. It turns out that the set of solutions of the equation $e^{2z} + 1 + 2z = 0$ contains a unique $\alpha$ that is closest to the origin $0$, and that $\alpha$ the unique real solution given by

$$ \alpha = -\frac{1 + W(e^{-1})}{2} \approx -0.63923227138053689755\cdots. $$

(Here, $W$ is the Lambert W-function.) Then we have

$$ (z - \alpha)f(z) = \alpha \sum_{n=0}^{\infty} (-1)^{n-1}(a_n + \alpha^{-1} a_{n-1}) z^n $$

Now the right-hand side has the radius of convergence which is strictly larger than $|\alpha|$, since it coincides with the modulus of the next closest zero of $e^{2z} + 1 + 2z = 0$. Thus for some $r \in (0, 1)$ we have

$$ a_n + \alpha^{-1}a_{n-1} = \mathcal{O}((r/\alpha)^n) \quad \Longrightarrow \quad a_n = \frac{C+\mathcal{O}(r^n)}{|\alpha|^n} $$

Plugging this to the ratio of the consecutive terms of $(\epsilon_n)$, we get

$$ \frac{\epsilon_{n}}{\epsilon_{n-1}} = \frac{n}{\alpha} + o(1). $$

Thus the number we are seeking is exactly

$$ \frac{1}{\alpha} \approx -1.5643765885603998024\cdots. $$

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  • $\begingroup$ What are the big O and little o notations you are using? $\endgroup$ – Eleven-Eleven Jun 11 '15 at 22:27
  • $\begingroup$ @Eleven-Eleven, Both are for $n \to \infty$. $\endgroup$ – Sangchul Lee Jun 11 '15 at 22:48
  • $\begingroup$ Okay. I thought so. $\endgroup$ – Eleven-Eleven Jun 12 '15 at 0:26

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