Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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-

share|cite|improve this question
Take the logarithm of both sides and use Taylor series. – Qiaochu Yuan Jul 25 '11 at 18:07

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.

share|cite|improve this answer

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.

share|cite|improve this answer
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

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.