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Interested by this question, $j$ being a positive integer, I tried to work the asymptotics of

$$S^{(j)}_n=\sum^{n}_{k=0}\frac{\binom{n}{k}}{n^k(k+j)}=\frac{\, _2F_1\left(j,-n;j+1;-\frac{1}{n}\right)}{j}$$

I quickly noticed (not a proof) that the asymptotics write

$$S^{(j)}_n=(-1)^j\left(\left(\alpha_0-\beta_0e\right)-\frac{\left(\alpha_1-\beta_1e\right)}{2n}+\frac{\left(\alpha_2-\beta_2e\right)}{24n^2}\right)+O\left(\frac{1}{n^3}\right)$$

in which the $\alpha_k$'s and $\beta_k$'s are all positive whole numbers depending on $j$.

What I found is that $$\alpha_0=(j-1)!\qquad \qquad \beta_0=\text{Subfactorial}[j-1]$$ $$\alpha_1=(j+1)!\qquad \qquad \beta_1=\text{Subfactorial}[j+1]$$ $$\alpha_2=(1+3j)(j+2)!$$ but I did not find any formal representation for $\beta_2$.

I give below a list of values ot this last coefficient as a function of $j$.

$$\left( \begin{array}{cc} j & \beta_2 \\ 1 & 11 \\ 2 & 60 \\ 3 & 443 \\ 4 & 3442 \\ 5 & 29667 \\ 6 & 281824 \\ 7 & 2936915 \\ 8 & 33374022 \\ 9 & 411167963 \\ 10 & 5462660068 \\ 11 & 77886959691 \\ 12 & 1186630738810 \\ 13 & 19242660629363 \\ 14 & 330973762825032 \end{array} \right)$$

It seems that $\frac{\beta_2}{\text{Subfactorial}[j+2]}$ is close to a straight line but it is not.

Is there any way to identify what is this sequence ?

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Claude Leibovici
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  • I checked the On-Line Encyclopedia of Integer Sequences, https://oeis.org/, and unfortunately it didn't know this sequence. – Imago Jul 11 '16 at 05:50
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    @Imago. Thanks but be sure I did ! In fact, OEIS gave me the very first terms. – Claude Leibovici Jul 11 '16 at 06:04
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    $\beta_2 = 3(!(j+3)) - 8(!(j+2))$ where $!n \stackrel{def}{=} n!\sum\limits_{k=0}^n \frac{(-1)^k}{k!} = \left\lfloor\frac{n!}{e} + \frac12\right\rfloor$ is the $n^{th}$ derangement number (aka subfactorial). – achille hui Jul 11 '16 at 06:42
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    @achillehui. Thank you so much ! This is perfect. May I ask how you did find it ? Why don't you post an answer ? – Claude Leibovici Jul 11 '16 at 07:02

1 Answers1

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Notice $$S_n^{(j)} = \int_0^1 \left(1+\frac{x}{n}\right)^n x^{j-1} dx = \int_0^1 e^{n\log\left(1+\frac{x}{n}\right)} x^{j-1} dx = \int_0^1 e^{x - \frac{x^2}{2n} + \frac{x^3}{3n^2} + O(n^{-3})} x^{j-1}dx\\ = \int_0^1 e^x \left[1 - \frac{x^2}{2n} + \frac{8x^3+3x^4}{24n^2} + O(n^{-3})\right]x^{j-1} dx $$ Compare with expansion in question, we get

$$(-1)^j(\alpha_2 - \beta_2 e) = 8A_{j+2}+3A_{j+3} $$ where $$\begin{align}A_{j} &= \int_0^1 e^x x^j dx = \left(\int_{-\infty}^1 - \int_{-\infty}^0\right) e^x x^j dx = \underbrace{\int_0^\infty e^{1-y} (1-y)^j dy}_{x=1-y} - \underbrace{\int_0^\infty e^{-y} (-y)^j dy}_{x = -y}\\ &= (-1)^j\left(e \int_0^\infty e^{-y} (y-1)^j dy - j!\right) \end{align} $$ Using the fact $$\int_0^\infty e^{-y} (y-1)^n dy = !n \stackrel{def}{=} n! \left(\sum_{s=0}^n \frac{(-1)^s}{s!}\right) = \left\lfloor\frac{n!}{e}+\frac12\right\rfloor$$ is the $n^{th}$ derangement number, we find $$(-1)^j\left(\alpha_2 - \beta_2 e\right) = (-1)^j\left\{8[e!(j+2) - (j+2)!] - 3[e!(j+3) - (j+3)!]\right\}\\ \implies \begin{cases} \alpha_2 &= 3\times(j+3)! - 8\times(j+2)! = (3j+1)(j+2)!\\ \beta_2 &= 3\,\times\,!(j+3) - 8\,\times\,!(j+2) \end{cases} $$

achille hui
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