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Short version: can we prove that $$\sum_{k=0}^n (-1)^k \binom{n}{k}^2 \frac{k!}{n^{2k}} \to \frac1e$$ as $n \to \infty$?

Long version: First, consider $$a_n = \sum_{k=0}^n \frac{(-1)^k}{k!}$$ It is well-known that $a_n \to \dfrac1e$ as $n \to \infty$; indeed $a_n = \dfrac{!n}{n!}$ is the truncation to the first $n$ terms of the power series for $e^x$, evaluated at $x = -1$ (where $!n$ denotes the subfactorial; it is also equal to the number of derangements on $n$ elements). There is also a simple expression for the generating function $A(z) = \sum_{n=0}^{\infty} a_n z^n$, which is $A(z) = \dfrac{e^{-z}}{1-z}$. (See Exponential Generating Functions For Derangements.)

Next, consider $$b_n = \sum_{k=0}^n \frac{n^{\underline k}}{n^k} \frac{(-1)^k}{k!}$$ where $n^{\underline k} = \binom{n}{k}k!$ denotes a falling factorial, so the extra factor $\frac{n^{\underline k}}{n^k}$ is $\frac{n(n-1)\cdots(n-k+1)}{n(n)\cdots(n)} = 1\cdot\left(1 - \frac1n\right)\cdot\left(1 - \frac2n\right)\cdots\left(1 - \frac{k-1}n\right)$ which for large $n$ (and fixed $k$) is close to $1$. I don't know if $b_n$ has a simple form for its generating function too, but it is easy to see that $b_n \to \dfrac1e$ as well; indeed by the binomial theorem we have $b_n = \sum_{k=0}^n \binom{n}{k} (\frac{-1}{n})^k = \left(1 - \frac1n \right)^n$ which is well-known to converge to $\frac1e$ (indeed such a limit is sometimes taken to be the definition of $e^x$).

Finally, consider $$c_n = \sum_{k=0}^n \frac{n^{\underline k}}{n^k} \frac{n^{\underline k}}{n^k} \frac{(-1)^k}{k!}$$

This is the same transformation going from $b_n$ to $c_n$ as from $a_n$ to $b_n$. But can we prove that $c_n \to \frac1e$ too? (And can we write down its generating function compactly, perchance?) More generally, what techniques exist that help in proving something about $\sum t_n s_n$, given $\sum s_n$?

This question arose from an attempt to answer this question, where I arrived at the expression $c_n$ above (there I called it $P_{n, n, 0}$; next I'll try to understand $P_{m, w, k}$).

[Note: I'm tagging this too, as I understand $c_n$ has something to do with hypergeometric functions / Bessel functions / something like that.]

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ShreevatsaR
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2 Answers2

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I hope I understood your question correctly.

For any sequence of real numbers $(\alpha_n)_{n\geq 0}$ and $p\in\mathbb N$ let us denote by $(\alpha_n^{*p})$ (awkward notation) the sequence defined by $$\alpha_n^{*p}=\sum_{k=0}^n \left(\frac{n^{\underline k}}{n^k}\right)^p \alpha_k\, . $$

Then, the following holds true: for any absolutely convergent series $\sum\alpha_k$, the sequence $(\alpha_n^{*p})_{n\geq 0}$ is convergent with limit $\sum_0^\infty\alpha_k$. In particular, with $\alpha_k=\frac{(-1)^k}{k!}$ and $p=2$ you get that $c_n\to1/e$.

To see this, note that one can write $$\alpha_n^{*p}=\sum_{k=0}^\infty q_{n,k}\, \alpha_k\, ,$$ where $$q_{n,k}=\left\{ \begin{matrix} \left(\frac{n^{\underline k}}{n^k}\right)^p&k\leq n\\0&k>n \end{matrix}\right. $$

For each fixed $k$ we have $\lim_{n\to\infty} q_{n,k}=1$ by the formula for $\frac{n^{\underline k}}{n^k}$ you give in your question. Moreover, since $0\leq q_{n,k}\leq 1$ by the same formula, we also have $\vert q_{n,k}\alpha_k\vert\leq \vert\alpha_k\vert$ for all $n$ and every $k\geq 0$. By the dominated convergence theorem (for series), it follows that $$\lim_{n\to\infty} \sum_{k=0}^\infty q_{n,k}\alpha_k=\sum_{k=0}^\infty\alpha_k\, , $$ which is the required result.

Etienne
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  • Thanks, "dominated convergence theorem (for series)" is what I was missing. (By the way, a minor note on simplifying notation: $n^{\underline k}$ turns out to be $0$ for $k > n$ (because there's a $(n-n)$ term in the falling factorial — alternatively note that $\binom{n}{k}k! = 0$ as $\binom{n}{k} = 0$ for $k > n$), so we don't really need a case-by-case definition of $q_{n,k}$. But perhaps it makes it clearer.) – ShreevatsaR Mar 26 '14 at 01:34
  • @ShreevatsaR Thanks for the note on the simplifying notation. Still, I think I'll leave the case-by-case definition. – Etienne Mar 26 '14 at 08:06
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(I'm answering my own question, to reflect what I've understood from user Etienne's answer.)

The relevant tool is the dominated convergence theorem for series (sums). There are references here and here, among others. It says that

if for every $k$, we have $f_n(k) \to f(k)$ as $n \to \infty$, and further each of the sequences $f_n$ is dominated by some summable sequence $g$ (that is $|f_n(k)| \le g(k)$ for all $n$ and $k$ where $\sum_{k} g(k)$ exists), then $$\lim_{n \to \infty} \sum_k f_n(k) = \sum_{k} \lim_{n\to\infty} f_n(k) = \sum_k f(k)$$

(I've intentionally left out the "types" of $k$, $f$, $f_n$, etc., to keep the statement concise and because the theorem is applicable broadly.)

In particular, suppose we know that a particular sequence $s_k$ has series sum $\sum_{k \ge 0} s_k = S$. Suppose further that $t_{m,k}$ is a sequence of tweaking/tempering factors, such that $|t_{m,k}| \le 1$, and $\lim_{m \to \infty} t_{m,k} = 1$ for all $k$. Then, $$\lim_{m\to\infty} \sum_{k \ge 0} t_{m,k} s_k = \sum_{k\ge0} \lim_{m\to\infty} t_{m,k} s_k = \sum_{k\ge 0} s_k = S.$$

Thus multiplying the $k$th term $s_k$ by $t_{m,k}$ (which in the limit is $1$) does not affect the sum in the limit.

What we have here is a particular case with $t_{m,k} = \dfrac{m^{\underline k}}{m^k}\dfrac{m^{\underline k}}{m^k}$, which clearly goes to $1$, and therefore the sum (for the particular $s_k$'s in the question) remains $1/e$.

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