# how to turn the sum $\lim\limits_{n\to\infty}\left(\prod\limits_{k=0}^n \frac{2n+2k}{2n+2k+1}\right)$ into integral with Riemann sum

I want to solve $$\lim_{n\to\infty}\left(\prod_{k=0}^n \frac{2n+2k}{2n+2k+1}\right)$$ with integration I know that $$\lim_{n \to \infty}\frac{b-a}{n} \sum_{k=0}^nf(a+\frac{b-a}{n})=\int_a^b f(x)dx$$ so let $$s=\lim_{n\to\infty}\left(\prod_{k=0}^n \frac{2n+2k}{2n+2k+1}\right)$$ then $$\ln(s)=\lim_{n\to\infty}\ln\left(\prod_{k=0}^n \frac{2n+2k}{2n+2k+1}\right)$$

$$-\ln(s)=\lim_{n\to\infty} \left(\sum_{k=0}^n \ln\left(1+\frac{1}{2n+2k}\right)\right)$$

and here I got stuck for an hour and I couldn't progress any further so I decided to ask here I couldn't "pull out" $$\frac{1}{n}$$ from the logarithm to turn the sum into Riemann sum
I also don't know what $$O(n)$$ means I see it very often in questions like this

• Apply $$x-{x^2\over 2}\le \ln(1+x)\le x$$ Commented Jul 12, 2023 at 23:22

$$\text{L}＝\lim\limits_{n\to\infty}\left(\prod\limits_{k=0}^n \frac{2n+2k}{2n+2k+1}\right)\\~\\-\ln(\text{L})＝\lim\limits_{n\to\infty}\left(\sum_{k＝0}^{n}\ln\left({1+\frac{1}{2n+2k}}\right)\right)\\~\\-\ln(\text{L})＝\lim\limits_{n\to\infty}\left(\sum_{k＝0}^{n}\ln\left({1+\frac {1}{n}\left({\frac {1}{2＋2\left({\frac {k}{n}}\right)}}\right)}\right)\right)\\~\\f(x)＝\frac {1}{2＋2x}\\~\\-\ln(\text{L})＝\lim\limits_{n\to\infty}\left(\sum_{k＝0}^{n}\ln\left({1+\frac {1}{n}f\left({\frac {k}{n}}\right)}\right)\right)\\~\\\;\ln(1＋x)＝x＋\mathcal{O}(x^2)\\~\\-\ln(\text{L})＝\lim\limits_{n\to\infty}\left(\sum_{k＝0}^{n}\frac {1}{n}f\left({\frac {k}{n}}\right)＋+\mathcal{O}\left(\frac{1}{n^2}\sum_{k=0}^n f\left(\frac{k}{n}\right)^2\right)\right)\\~\\-\ln(\text{L})＝\lim\limits_{n\to\infty}\frac {1}{n}\sum_{k＝0}^{n}f\left({\frac {k}{n}}\right)＝\int_{0}^{1}f(x)\;dx\\~\\-\ln(\text{L})＝\int_{0}^{1}\;\frac {dx}{2＋2x}＝\ln(\sqrt {2})\\~\\\text{L}＝\lim\limits_{n\to\infty}\left(\prod\limits_{k=0}^n \frac{2n+2k}{2n+2k+1}\right)＝\frac {1}{\sqrt {2}}$$

I figure out another way to solve this I am not sure if it is right or wrong

let $$L=\lim_{n\to\infty}\left(\prod_{k=0}^n \frac{2n+2k}{2n+2k+1}\right)$$ $$\ln(L)=\lim_{n\to\infty}\left(\sum_{k=0}^n \ln(2n+2k)\right)-\lim_{n\to\infty}\left(\sum_{k=0}^n \ln(2n+2k+1)\right)$$ let $$a=\frac{1}{n}$$

then $$\ln(L)=\lim_{n\to\infty}\left(\sum_{k=0}^n\ln(n)+\sum_{k=0}^n \ln(2n+2ak)-\sum_{k=0}^n\ln(n)-\sum_{k=0}^n \ln(2n+2a+a)\right)$$ $$\ln(L)=\lim_{a\to0}\left(\sum\limits_{k=0}^n \ln(2+2ka)-\sum_{k=0}^n \ln(2+2ka+a)\right)$$ $$\ln(L)=\lim_{a\to0}-a\frac{\left(\sum\limits_{k=0}^n- \ln(2+2ka)+ \ln(2+2ka+a)\right)}{a}$$ $$m:=\lim_{a\to0}\frac{- \ln(2+2ka)+ \ln(2+2ka+a)}{a}$$is the defintion of derivative

then $$\ln(L)=-\lim_{a\to0}a\sum\limits_{k=0}^n\frac{1}{2+2ka}$$ which is Riemann sum so $$\ln(L)=\int_0^1\frac{-1}{2+2x}dx$$ $$\ln(L)=-0.5\ln(2)$$ $$\ln(L)=\ln(\frac{1}{\sqrt{2}})$$ $$L=\frac{1}{\sqrt{2}}$$

• why did you flip your final result? Commented Jul 15, 2023 at 11:24
• it was an error and i fixed it
– pie
Commented Jul 15, 2023 at 11:25