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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.

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Note: the product is similar in form to the generalized binomial coefficient: en.wikipedia.org/wiki/… . I don't know if it helps. –  Erel Segal Halevi Jun 27 '13 at 6:53
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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
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(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

1 Answer 1

up vote 2 down vote accepted

We have:

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

for all $x$.

So:

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

where:

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

Hence:

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

as $H_n \to \infty.$

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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

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