Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The sum is:

$$ S = 1 + 1/2 + \frac {(n-1)(n-2)} {3n^2} + \frac {(n-1)(n-2)(n-3)} {4n^3} + \ldots + \frac {(n-1)!} {n \times n^{n-1}}$$ $$= \frac 3 2 + \sum_{k=3}^{n} \frac{n!}{k(n-k)!n^k} $$

Can we get an asymptotic lower bound of $S$?

I guess it's $\Omega(\frac 1 2 \log n)$, but I'm not sure how to get it. $k$ starts from 3 because I'm actually counting the expectation of cycles in a graph. And least 3 nodes could form a cycle. But since what I need is an asymptotic lower bound, I guess where $k$ starts doesn't really matter.

share|cite|improve this question
The first two terms don't seem to fit a simple pattern, and why are there two $(n-2)$ factors in second numerator? Is the general pattern supposed to be something like $S(n)=\sum_{k=1}^{n} \frac{n!}{k(n-k)!n^k}$ ? – Henning Makholm Oct 10 '11 at 2:19
Why isn't the second term $(n-1)/(2n)$? Should the fourth term have two $(n-2)$'s or is the second of these supposed to be $(n-3)$? Lastly, I think one should write $S(n)$ since it's a function of $n$ (and they aren't exactly partial sums because the terms themselves vary with $n$). – anon Oct 10 '11 at 2:20
If this is $S(n) =\sum_{k=1}^{n-1} \frac{n!}{(n-k)!k n^k}$ the second term should be $\frac{n-1}{2n}$, not $\frac{1}{2}$, and the fourth should be $\frac{(n-1)(n-2)(n-3)}{4n^4}$. – Robert Israel Oct 10 '11 at 2:36
The 2 $(n-2)$ is a typo, I just fixed it. – ablmf Oct 10 '11 at 2:40
up vote 3 down vote accepted

The term $a_k(n) = \frac{1}{k} \prod_{j=1}^{k-1} (1 - j/n)$ satisfies $a_k(n) < 1/k$, so $S(n) < 3/2 + \sum_{k=3}^n 1/k \approx \ln(n)$. On the other hand, take any $p$ with $0 < p < 1/2$. For $k < n^p$ we have $\prod_{j=1}^{k-1} (1 - j/n) > (1 - n^{p-1})^{n^{p}}$, which has limit 1 as $n \to \infty$. So for any $\epsilon > 0$, if $n$ is sufficiently large we have $$S(n) > \sum_{k=3}^{n^{p}} \frac{1-\epsilon}{k} \approx (1-\epsilon) \ln(n^p) = (1-\epsilon) p \ln(n)$$

share|cite|improve this answer
That's really impressive. – ablmf Oct 10 '11 at 11:49

For a complete asymptotic up to $o(1)$ I get $$S(n) = \frac{1}{2} \log n + \frac{\gamma + \log 2}{2} + o(1).$$

Your sum $S(n)$ is very close to the sum $Q(n) = 1 + \frac{n-1}{n} + \frac{(n-1)(n-2)}{n^2} + \cdots = \sum_{k \geq 1} \frac{n^{\underline{k}}}{n^k}$ considered in Problem 9.56 in Concrete Mathematics and elsewhere in Knuth's work, so I've adapted some of the arguments I found there.

Let's consider the sum $S'(n) = \sum_{k \geq 1} \frac{n^{\underline{k}}}{k n^k} = S(n) - \frac{1}{2n} = S(n) + o(1).$ In the answer to Problem 9.56 in Concrete Mathematics the authors indicate that Stirling's approximation can be used to show that if $k \leq n^{1/2+\epsilon}$ then

$$\frac{n^{\underline{k}}}{k n^k} = e^{-k^2/2n} \left(\frac{1}{k} + \frac{1}{2n} - \frac{2}{3} \frac{k^2}{(2n)^2} + O(n^{-1+4 \epsilon})\right).$$ Then, Knuth and Pittel, in "A Recurrence Related to Trees," (Proceedings of the AMS 105(2) 1989, pp. 335-349) indicate this means that $\frac{n^{\underline{k}}}{k \, n^k}$ is exponentially small when $k \geq n^{1/2+\epsilon}$ and so can be replaced with other exponentially small terms to get $$S'(n) = T_{2n}(-1) + \left(\frac{1}{2n} + O(n^{-1 + 4 \epsilon})\right) T_{2n}(0) - \frac{1}{6n^2} T_{2n}(2),$$ where $T_n(x) = \sum_{k \geq 1} k^x e^{-k^2/n}$.

Lemma 1 in the Knuth and Pittel paper then states that if $x > -1$ then $$T_n(x) = \frac{1}{2} \Gamma\left(\frac{x+1}{2}\right) n^{(x+1)/2} + O(1).$$ They also mention that a derivation of $$T_n(-1) = \frac{1}{2} \log n + \frac{\gamma}{2} + O(n^{-1})$$ is in Knuth's Art of Computer Programming, Vol. 3, Exercise 5.2.2-4, as part of the analysis of bubblesort.

Putting this all together gives us $S(n) = \frac{1}{2} \log (2n) + \frac{\gamma}{2} + o(1) = \frac{1}{2} \log n + \frac{\gamma + \log 2}{2} + o(1).$

For more on the $Q(n)$ and related functions and their asymptotics, see The Art of Computer Programming, Vol. 1 (3rd ed.), Section

share|cite|improve this answer
I did computer simulation and $S(n)$ indeed goes to $1/2 log(n)$. But I am afraid your proof is a difficult for me to understand. Is there any other simpler proofs? – ablmf Oct 12 '11 at 23:57
@ablmf: Unfortunately, I don't know of a simpler one. I see that you have asked for a simpler proof as a separate question, though, so maybe somebody else on the site can think of one. However, if Knuth and Pittel felt the need to go through work like this to solve a similar problem, my guess is that finding a simpler proof won't be easy. – Mike Spivey Oct 13 '11 at 1:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.