Find $\lim\limits_{n \to \infty} \frac{1!+3!+\ldots+(2n-1)!}{2!+4!+\ldots+(2n)!}$

Question

$$\lim\limits_{n \to \infty} \frac{1!+3!+\ldots+(2n-1)!}{2!+4!+\ldots+(2n)!}$$

I have tried some standard approaches like dividing by $(2n)!$ and comparing consecutive terms. Hint, please :) Thanks in advance.

Answer 1

Morally: The last term in the denominator is $n$ times larger than the last term in the numerator, and all lower terms are negligible since they're smaller by a factor on the order $n^2$. This kills the limit. There is a minor technicality to be overcome, since there are still $n$ terms in the numerator.

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This can be estimated very brutally. The bottom is at least $(2n)!$, while the numerator is at most $(2n - 1)! + n \cdot (2n - 3)!$. This leads to

\begin{align*} \frac{(2n - 1)! + n (2n - 3)!}{(2n)!} &= \frac{\big((2n - 1)(2n - 2) + n\big)(2n - 3)!}{(2n)(2n - 1)(2n - 2)(2n - 3)!} \\ &= \frac{4n^2 + O(n)}{8n^3 + O(n^2)} \to 0 \end{align*}