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

What is the limit, when $n$ goes to $\infty$, of the following product, when $0 \leq a \leq 1$?

$$ {{1-a} \over 1}\cdot {{2-a} \over 2} \cdot {{3-a} \over 3} \cdot\ldots\cdot {{n-a} \over n} $$

When $a=0$, the product is 1, and when $a=1$, the product is 0, so I assume the product decreases monotonically with $a$ (actually, from the first factor it is clear that the product is always at most $1-a$). But I could not find any better approximations.

share|cite|improve this question
Note: the product is similar in form to the generalized binomial coefficient:… . I don't know if it helps. – Erel Segal-Halevi Jun 27 '13 at 6:53
Here is a useful general fact: let $\prod (1 + a_k)$ be an infinite product such that $\sum a_k^2$ converges. Then the infinite product converges iff $\sum a_k$ converges. You can prove this by taking logarithms. – Qiaochu Yuan Jun 27 '13 at 19:09
(Here I am using a notion of convergence of an infinite product where convergence to $0$ counts as a divergence, since after taking logarithms it corresponds to diverging to $-\infty$.) – Qiaochu Yuan Jun 27 '13 at 19:12
up vote 2 down vote accepted

We have:

$$\exp x \geq 1 + x $$

for all $x$.


$$ 0 \leq \prod_{k=1}^{n} (1 - {a \over k} ) \leq \exp (-a H_n) $$


$$H_n = \sum_{k=1}^{n} \frac{1}{k}$$


$$\lim_{n\to\infty} \prod_{k=1}^{n} (1 - {a \over k} ) = 0 $$

as $H_n \to \infty.$

share|cite|improve this answer
Isn't it true that $H_n < 1+ln(n)$? If so, then the upper bound is at most $exp(-a)/(n^a)$. – Erel Segal-Halevi Jun 28 '13 at 12:38

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.