I want to evaluate the following limit:
$$ \lim_{n\to \infty} \frac{3n+1}{3n}\frac{3n+2019+1}{3n+2019}...\frac{3n+2019n+1}{3n+2029n} $$
The general term of the product is: $$ \frac{3n+2019k+1}{3n+2019k}$$ Its limit seems to be one. Writing the initial limit as: $$ \lim_{n\to \infty} \prod_{k=0}^n\frac{(3n+2019k+1)}{(3n+2019k)} $$ I concluded that the limit would be 1. To my surprise, the answer seems to be $ \sqrt[2019]{674} $, which is very close to 1, though not the same thing.
This question was part of a highschool grade 12 competition. Can someone explain what my mistake is? Was I correct when reversing the limit and the product?
Thank you!
We have:
$$\lim_{n\to \infty} \prod_{k=0}^n\left(1+\frac{1}{3n+2019k}\right)=\lim_{n\to \infty}\operatorname{exp}\left(\sum_{k=0}^n\ln\left(1+\frac{1}{3n+2019k}\right)\right)$$
So we evaluate the limit without the exponential:
$$ \begin{aligned}\lim_{n\to \infty}\sum_{k=0}^n\ln\left(1+\frac{1}{3n+2019k}\right)&= \lim_{n\to \infty}\sum_{k=0}^n\frac{1}{3n+2019k}\\ &= \frac{1}{n}\lim_{n\to \infty}\sum_{k=0}^n\frac{1}{3+2019\frac{k}{n}}\\ &= \int_{0}^1\frac{1}{3+2019x}\,dx\\ &= \frac{1}{2019}\ln 674 \end{aligned} $$
and from the exponential you get the result $\sqrt[2019]{674}$.
Edit: The step where I pass from $\ln\left(1+\dfrac{1}{3n+2019k}\right)$ to $\dfrac{1}{3n+2019k}$ is based on:
$$x-x^2\leq \ln(1+x)\leq x$$
summing, and squeezing.