So I have to compute the following limit: $$\lim_{n \to \infty}{ \prod_{k = 0}^{n} \frac{3n + 2019k + 1}{3n + 2019k} }.$$ My first stab at it was to rewrite the limit as $$e^{\lim\limits_{n \to \infty} \ln\left( \prod_{k = 0}^{n} \frac{3n + 2019k + 1}{3n + 2019k} \right) }\;,$$ in order to arrive at an infinite sum: $$e^{\lim\limits_{n \to \infty}{ \sum_{k = 0}^{n} \ln\left(\frac{3n + 2019k + 1}{3n + 2019k}\right) }}\;.$$ I was hoping I could maybe turn it into a definite integral or something, but I'm not sure how I would do that having reached this point. Any help would be greatly appreciated.
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It is technically not an infinite sum. $$\lim_{n\to\infty}\sum_{k=1}^n f(n,k)$$ is a limit of sums, but not an infinite sum, because as $n$ increases, the terms change everywhere, not just adding new terms to the end. – Thomas Andrews May 04 '23 at 16:37
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Write you terms as $1+\frac1{3n+2019k}$ and use the power series for $\log(1+x).$ – Thomas Andrews May 04 '23 at 16:40
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Mathematica answers $\sqrt[2019]{674} $. – user64494 May 04 '23 at 16:44
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Rewrite your terms as $1+\frac1{3n+2019k},$ then use, for $|x|<1,$ that $\log(1+x)=x+O(x^2).$
So your logarithm is:
$$\sum_{k=1}^n \left(\frac1{3n+2019k}+O(\frac1{n^2})\right)=O(\frac1n)+\sum_{k=1}^{n}\frac1{3n+2019k}$$
So you need to find the limit of $$\sum_{k=1}^n\frac1{3n+2019k}.\tag1$$
You can see $(1)$ is a Riemann sum for:
$$L=\int_{0}^{1}\frac{1}{3+2019x}\,dx,$$
then take $e^L$ for your limit.
Thomas Andrews
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