product converging to exp(x+1/2) 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-
 A: 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. 
A: 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. 
