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Consider the series

$$\sum\limits_{k=0}^\infty (-1)^kk!x^k\in\mathbb{R}[[x]]$$

Let $s_n(x)$ be partial sum. And let $\omega_{k,n}=(k!^2(n-k)!)^{-1}$. My question is:

Prove that $$\lim\limits_{n\to\infty}\dfrac{s_0\omega_{0,n}+\cdots + s_n\omega_{n,n} }{\omega_{0,n}+\dots+\omega_{n,n}}=\int\limits_0^\infty\dfrac{e^{-t}}{xt+1}dt$$

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an interesting question! (+1) – user 1618033 Aug 11 '12 at 12:13
With no motivation whatsoever, nor any hint about what the OP tried, or why the answer should be such. – Did Aug 11 '12 at 22:03
$$\int\limits_0^{\infty}\dfrac{e^{-t}}{xt+1}dt=s_n(x)+o(x^n)$$ – aska langley Aug 11 '12 at 22:25
Numerical calculations in maple: for $x=0.7, n=10$ left-hand side is 0.6634301660, right-hand side is 0.6635102741. – aska langley Aug 11 '12 at 22:32
Come on! What are you trying to make us believe? This is certainly not by numerical calculations that one can guess the formula on the LHS. – Did Aug 11 '12 at 22:37
up vote 4 down vote accepted

Since did already said a lot, let me just offer a few more observations that are too long for a comment:

The limit of the weighted mean summation (i.e. somewhat akin to Cesàro summation, but with weights) being considered might look a bit less forbidding if rewritten in terms of forward differences:

$$A_n=\frac{s_0\omega_{0,n}+\cdots + s_n\omega_{n,n} }{\omega_{0,n}+\dots+\omega_{n,n}}=\left.\frac{\Delta^n\left(\frac{s_k}{(-1)^k k!}\right)}{\Delta^n\left(\frac1{(-1)^k k!}\right)}\right|_{k=0}$$

In fact, what you are looking at is a special case of a family of extrapolation algorithms studied by Avram Sidi (see e.g. this or this or this). In those references, the relation between Borel summation (which is the usual method applied for summing divergent series like the one in the OP) and these extrapolation methods is noted. An efficient recursive algorithm for computing these extrapolations is also given.

On the other hand, the convergence of the OP's particular special case is not too good, compared to some other cases of the general extrapolation method, e.g.

$$\begin{align*} \mathcal D_n&=\left.\frac{\Delta^n\left(\frac{s_k}{(-1)^k k! x^k}\right)}{\Delta^n\left(\frac1{(-1)^k k! x^k}\right)}\right|_{k=0}\\ \mathcal L_n&=\left.\frac{\Delta^n\left(\frac{s_k (k+1)^{n-1}}{(-1)^k k! x^k}\right)}{\Delta^n\left(\frac{(k+1)^{n-1}}{(-1)^k k! x^k}\right)}\right|_{k=0}\\ \mathcal S_n&=\left.\frac{\Delta^n\left(\frac{s_k (k+1)_{n-1}}{(-1)^k k! x^k}\right)}{\Delta^n\left(\frac{(k+1)_{n-1}}{(-1)^k k! x^k}\right)}\right|_{k=0} \end{align*}$$

where $(a)_k$ is the Pochhammer symbol. In particular, the transformation $\mathcal D_n$ is a convergence acceleration method studied by Drummond, while $\mathcal L_n$ is in fact the Levin $t$-transformation (see this related question), and $\mathcal S_n$ is the modification of the Levin transformation due to E.J. Weniger.

Here's a short table comparing the numerical performance of the various transformations with the regularized sum, $F(x)=\frac1{x}\exp\left(\frac1{x}\right)E_1\left(\frac1{x}\right)$ ($E_1(x)$ is the exponential integral), taking $n=10$:

\begin{array}{c|cc}x&F(x)&A_{10}&\mathcal D_{10}&\mathcal L_{10}&\mathcal S_{10}\\\hline 0.1&0.915633339&0.915625503&0.915633339&0.915633339&0.915633339\\ 0.2&0.852110881&0.851748876&0.852110881&0.852110881&0.852110881\\ 0.5&0.722657234&0.722476442&0.722656381&0.722657245&0.722657234\\ 0.75&0.650812854&0.650744812&0.650803895&0.650812841&0.650812848\\ 1&0.596347362&0.596310789&0.596310789&0.596347200&0.596347353\\ 1.5&0.517329839&0.517327162&0.517138593&0.517329576&0.517329988\\ 2&0.461455316&0.476549262&0.460953049&0.461456123&0.461455825\\ \end{array}

Larger values of $n$ will give better results, up to a point, when subtractive cancellation from subtracting large terms sets in.

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  • For every fixed $i$, identify the coefficient of $x^i$ in the LHS and in the RHS.
  • Deduce that the task is to prove that $\lim\limits_{n\to\infty}\dfrac{\omega_{i,n}+\cdots+\omega_{n,n}}{\omega_{0,n}+\cdots+\omega_{n,n}}=1$, for every fixed $i$.
  • Equivalently, one must show that $\lim\limits_{n\to\infty}\dfrac{\omega_{k,n}}{\omega_{0,n}+\cdots+\omega_{n,n}}=0$, for every fixed $k$.
  • Deduce that, if $\lim\limits_{n\to\infty}\dfrac{\omega_{k,n}}{\omega_{k+1,n}}=0$, for every fixed $k$, then the result holds.
  • Compute the ratio $\dfrac{\omega_{k,n}}{\omega_{k+1,n}}$ and its limit when $n\to\infty$, and conclude.
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"For every fixed i, identify the coefficient of xi in the LHS and in the RHS." - How can this be used? – aska langley Aug 11 '12 at 22:38
Did you identify these? – Did Aug 11 '12 at 22:39
aska: I love your reaction to my two comments, here and on the main question. – Did Aug 12 '12 at 13:25

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