## Find the limit of width and height ratio

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### Solution

For the geometric construction, we have $$L_{0} H_{0}=1, L_{1} H_{0}=2, L_{1} H_{1}=3$$ so more in general $$L_{i} H_{j}=1+i+j$$. Around the nth step the equations are $\left\{\begin{array}{c} L_{n} H_{n-1}=2 n \\ L_{n} H_{n}=2 n+1 \\ L_{n+1} H_{n}=2 n+2 \end{array}\right.$
Two recurrence sequences can be obtained by dividing the second equation with the first one and the third equation with the second one $\left\{\begin{array} { c } { H _ { 0 } = 1 } \\ { H _ { n } = H _ { n – 1 } \frac { 2 n + 1 } { 2 n } } \end{array} \quad \left\{\begin{array}{c} L_{0}=1 \\ L_{n}=L_{n-1} \frac{2 n+2}{2 n+1} \end{array}\right.\right.$ Explaining each recurrence sequence $H_{n}=\frac{2 n+1}{2 n} H_{n-1}$ $=\frac{2 n+1}{2 n} \frac{2 n-1}{2 n-2} H_{n-2}$ $\vdots$ $=\frac{(2 n+1)(2 n+1) \cdots 3}{2 n(2 n-2) \cdots 2} H_{0}$ $=\prod_{k=1}^{n} \frac{2 k+1}{2 k}$ $=\frac{(2 n+1) ! !}{2^{n} n !}$ $=\frac{(2 n+1)(2 n) !}{4^{n}(n !)^{2}}$ $L_{n}=\frac{2 n}{2 n-1} L_{n-1}$ $=\frac{2 n}{2 n-1} \frac{2 n-2}{2 n-3} L_{n-2}$ $\vdots$ $=\frac{2 n(2 n-2) \cdots 2}{(2 n-1)(2 n-3) \cdots 1} L_{0}$ $=\prod_{k=1}^{n} \frac{2 k}{2 k-1}$ $=\frac{2^{n} n !}{(2 n-1) ! !}$ $=\frac{4^{n}(n !)^{2}}{(2 n) !}$
So $\lim _{n \rightarrow \infty} \frac{L_{n}}{H_{n}}=\lim _{n \rightarrow \infty} \frac{16^{n}(n !)^{4}}{[(2 n) !]^{2}(2 n+1)}$
$\text { Using Stirling’s formula }$ $\left(n ! \approx n^{n} e^{-n} \sqrt{2 \pi n} \text { if } n\right. \text { goes to infinity) }$ $(n !)^{4}=n^{4 n} e^{-4 n} 4 \pi^{2} n^{2}$ $\quad[(2 n) !]^{2}=\left[(2 n)^{2 n} e^{-2 n} \sqrt{4 \pi n}\right]^{2}=16^{n} e^{-4 n} 4 \pi n$
$\large \lim _{n \rightarrow \infty} \frac{L_{n}}{H_{n}}=\lim _{n \rightarrow \infty} \frac{\pi n}{2 n+1}=\frac{\pi}{2}$
Credit: Giuseppe Palaia

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