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You're right. It is easy with Stirling's approximation $$N! \sim \sqrt{2\pi N}\left(\frac{N}e\right)^N$$ so that $$L_N=\frac{(2N)!}{N!N!2^{2N}} \sim \frac{\sqrt{2\pi(2N)}(2N)^{2N}}{e^{2N}2^{2N}}\frac …

Start by re-expressing the product term \begin{align}\frac{(kn-k)(kn-k-1)\cdots(kn-k-i)}{(kn)(kn-1)\cdots(kn-i)} &=\frac{(kn-i-1)\cdots(kn-i-k)}{(kn)\cdots(kn-k+1)}=\frac{\binom{kn-1-i}k}{\binom{kn}k} …

Here is an explicit formula for your ratio $r_n=\frac{n_n}{d_n}$: $$r_n= \frac{\sum_{k=0}^n\binom{n+k}{2k}(-x)^k} {\sum_{k=0}^n\binom{n+k+1}{2k+1}(-x)^k}.$$ Let $P_n(x)$ and $Q_n(x)$ be the numerator …

Let $g(a)=\frac{a}{1-a}, f(a)=\log(1-a)$ and $h(a)=\log^2(1-a)$. Let $\square^m$ denote a $m$-dimensional unit hypercube. The following is an application or reformulation of the above infinite series …

This is just a note. There is a Riemann sum involved here: $$\lim_{n\rightarrow\infty}R_n(z):=\lim_{n\rightarrow\infty}\,\,\frac1n\sum_{j=1}^n\sqrt\frac{n}{j}\sin\left(z\log{\frac{n}{j}}\right) =-\i …

Splitting your sum as $\sum_{t=0}^{n-1}=1+\sum_{t=1}^{n-1}$, it suffices to show that $\sum_{t=1}^{n-1}\binom{n-1}ta^{t(n-t)}\rightarrow0$ as $n\rightarrow\infty$; provided $0<a<1$. To illustrate our …