I'm having trouble calculating this limit directly :
$$\lim_{n\to\infty}\frac{(2n＋1)(2n＋3)\cdots(4n＋1)}{(2n)(2n＋2)\cdots(4n)}$$
It can be calculated using the inventory method and the result is:
$$\lim_{n\to\infty}\frac{(2n＋1)(2n＋3)\cdots(4n＋1)}{(2n)(2n＋2)\cdots(4n)} ＝\sqrt {2}$$
But the limit method is not smart, I need help doing this in a direct way.
This is the first time I am asking a question on this site. I apologize if it is not a good question.
You ask: what is the limit of
$$\lim_{n\to\infty}\prod_{k=n}^{2n} \frac{2k+1}{2k}$$
Note that
$$\frac{2k+1}{2k}=1+\frac{1}{2k}>1$$
Thus, we can write
$$\lim_{n\to\infty}\prod_{k=n}^{2n} \frac{2k+1}{2k}=\exp\left(\ln\left(\lim_{n\to\infty}\prod_{k=n}^{2n} \frac{2k+1}{2k}\right)\right)=\exp\left(\lim_{n\to\infty}\sum_{k=n}^{2n} \ln\left(\frac{2k+1}{2k}\right)\right)$$
This is allowed since the limit, IF IT EXISTS, is certainly greater than or equal to $1$. So the question now becomes, what is
$$\lim_{n\to\infty}\sum_{k=n}^{2n} \ln\left(1+\frac{1}{2k}\right)$$
Define
$$f(n)=\sum_{k=1}^{2n} \ln\left(1+\frac{1}{2k}\right)$$
Then our limit question becomes: what is $\lim_{n\to\infty} [f(2n)-f(n-1)]$?
We can expand the log term as
$$\ln\left(1+\frac{1}{2k}\right)=-\sum_{r=1}^\infty \frac{(-1)^r}{r(2k)^r}=\frac{1}{2k}-\sum_{r=2}^\infty \frac{(-1)^r}{r(2k)^r}$$
This gives us that $f(n)$ is
$$f(n)=\sum_{k=1}^n \left[\frac{1}{2k}-\sum_{r=2}^\infty \frac{(-1)^r}{r(2k)^r}\right]$$
Switching the order of sums here here (which is allowed since $2k>1$ implies the sum converges absolutely) then gives us
$$=\frac{1}{2}\sum_{k=1}^n \frac{1}{k}-\sum_{r=2}^\infty\frac{(-1)^r}{r2^r}\sum_{k=1}^n\frac{1}{k^r}=\frac{H_n}{2}-\sum_{r=2}^\infty\frac{(-1)^r}{r2^r}\sum_{k=1}^n\frac{1}{k^r}$$
where $H_n$ is the $n$ Harmonic number. Then using this formula for $f(n)$ we have
$$\lim_{n\to\infty} [f(2n)-f(n-1)]=\lim_{n\to\infty}\left[ \frac{H_{2n}}{2}-\sum_{r=2}^\infty\frac{(-1)^r}{r2^r}\sum_{k=1}^{2n}\frac{1}{k^r}-\frac{H_{n-1}}{2}+\sum_{r=2}^\infty\frac{(-1)^r}{r2^r}\sum_{k=1}^{n-1}\frac{1}{k^r}\right]$$
Now, two of these sums converge: we have
$$\lim_{n\to\infty} \sum_{k=1}^n \frac{1}{k^r}=\zeta(r)$$
for $r>1$. Then
$$=\lim_{n\to\infty}\left[ \frac{H_{2n}}{2}-\sum_{r=2}^\infty\frac{(-1)^r}{r2^r}\zeta(r)-\frac{H_{n-1}}{2}+\sum_{r=2}^\infty\frac{(-1)^r}{r2^r}\zeta(r)\right]=\lim_{n\to\infty}\frac{H_{2n}-H_{n-1}}{2}$$
For conveniences sake, note that
$$\lim_{n\to\infty}[H_n-H_{n-1}]=\lim_{n\to\infty}\frac{1}{n}=0$$
Thus, the question becomes: what is
$$\lim_{n\to\infty}[H_{2n}-H_n]$$
This has many methods/answers already on this site, but choosing any of them gives us that
$$\lim_{n\to\infty}[H_{2n}-H_n]=\ln(2)$$
Putting it all together, we have
$$\lim_{n\to\infty}\prod_{k=n}^{2n} \frac{2k+1}{2k}=\exp\left(\frac{\ln(2)}{2}\right)=\sqrt{2}$$
I don't know what are inventory method or limitation method, but I would solve it this way:
$$\frac{(2n+1)\cdot\ldots\cdot(4n+1)}{(2n)\cdot\ldots\cdot(4n)} = \frac{(2n)\cdot(2n+1)\cdot\ldots\cdot(4n)\cdot(4n+1)}{((2n)\cdot\ldots\cdot(4n))^2}$$
Now, the numerator is equal to $(4n+1)!$ divided by $(2n-1)!$, while the denominator is
$$((2n)\cdot(2n+2)\cdot\ldots\cdot(4n))^2=(2^{n+1})^2\cdot((n)\cdot(n+1)\cdot\ldots\cdot(2n))^2=(2^{n+1}\cdot (2n)!\cdot((n-1)!)^{-1})^2$$
so you get that your limit is equal to
$$\lim_{n→+\infty} \frac{(4n+1)!\cdot(n-1)!^2}{2^{2n+2}\cdot(2n)!^2\cdot(2n-1)!}$$
Now I would use Stirling's approximation formula $k! \sim \sqrt{2\pi k}\cdot k^k\cdot e^{-k}$ and algebraic manipulations to get the thesis.
