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Sorry if this is an easy one- I've tried everything I can think of (which isn't much), but hopefully there's an easy answer I'm not getting.

I'd like to show that $$\lim_{n\to \infty} \prod_{k=1}^n \left(1+\frac an + \frac {bk}{n^2}\right) = e^{a + b/2}.$$ Wolfram Alpha says it's true (for particular values of a and b), but doesn't say why. I guess you could expand the product by hand and just look at the part that's constant in $n$, but this immediately gets very complicated and I'm not used to making counts like that. Any other ideas? Thanks a lot-

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Take the logarithm of both sides and use Taylor series. –  Qiaochu Yuan Jul 25 '11 at 18:07

2 Answers 2

Here is the intuition:

Since $\sum_{k=1}^n =\frac{n(n+1)}{2}\sim \frac{n^2}{2}$, we expect that $$\prod_{k=1}^n \left(1+\frac an + \frac {bk}{n^2}\right) \sim \left(1+\frac an + \frac {b}{2n}\right)^n.$$ But we know that $$\lim_{n\to \infty} \left(1+\frac an + \frac {b}{2n}\right)^n =e^{a+\frac{b}{2}}.$$

Using logarithms, you can formalize this intuition.

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First hint. Use the inequality: $e^{x - x^2} \leq 1+x \leq e^{x}$ valid for $|x| \leq 1/4$.

Full Solution. As a first step, I'll use the hint to get the inequality @George suggests: $$ 1+\frac{a}{n}+\frac{bk}{n^2} \leq \exp(\frac{a}{n}+\frac{bk}{n^2}), $$ and $$ 1+\frac{a}{n}+\frac{bk}{n^2} \geq \exp(\frac{a}{n}+\frac{bk}{n^2} - (\frac{a}{n}+\frac{bk}{n^2})^2) \geq \exp(\frac{a}{n}+\frac{bk}{n^2} - \frac{(a+b)^2}{n^2}). $$

Now we will use this to get the limit. We have the following upper bound on the product: $$ \prod_{k=1}^n (1+\frac{a}{n}+\frac{bk}{n^2}) \leq \prod_{k=1}^{n} \exp(\frac{a}{n}+\frac{bk}{n^2}) = \exp(\frac{a}{n}n + \frac{b}{n^2} \frac{n(n+1)}{2}) = \exp(a+\frac{b}{2}+\frac{b}{2n}). $$ We can also establish a similar lower bound: $$ \prod_{k=1}^n (1+\frac{a}{n}+\frac{bk}{n^2}) \geq \prod_{k=1}^{n} \exp(\frac{a}{n}+\frac{bk}{n^2} - \frac{(a+b)^2}{n^2}) = \exp(a + \frac{b}{2} + \frac{b}{2n} - \frac{(a+b)^2}{n}). $$ Note that both the upper and lower bounds approach the limit $\exp(a+\frac{b}{2})$ as $n \to \infty$. Therefore, by the squeeze theorem, the given product also has the same limit.

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Or you can use $\log(1+a/n+bk/n^2)=a/n+bk/n^2+O(1/n^2)$, which amounts to the same thing. –  George Lowther Jul 25 '11 at 18:11

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