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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 and denominator polynomials of $r_n$, respectively. Then both polynomials share a common recurrence; namely, $$P_{n+2}+(x-2)P_{n+1}+P_n=0 \qquad \text{and} \qquad Q_{n+2}+(x-2)Q_{n+1}+Q_n=0.\tag1$$ They differ only in the initial condition where $P_0=1, P_1=1-x$ while $Q_0=1, Q_1=2-x$. The importance of such a description is that (1) the original recursive relations are decoupled here; (2) it is more amenable to an asymptotic analysis; (3) it reveals the roots being in $[0,4]$ due to the interlacing property of three-term recurrences.

Note. The original numerator and denominator differ by $\pm$ sign from $P_n$ and $Q_n$, but this makes no difference for the ratio $r_n$.

In (1), multiply the 1st equation through by $Q_{n+2}$, the 2nd by $P_{n+2}$ and subtract to get $$(x-2)[P_{n+1}Q_{n+2}-Q_{n+1}P_{n+2}]=Q_nP_{n+2}-P_nQ_{n+2}.$$ Divide by $Q_{n+1}Q_{n+2}$ and rearrange a bit to get $$(x-2)\left(\frac{P_{n+1}}{Q_{n+1}}-\frac{P_{n+2}}{Q_{n+2}}\right) =\frac{Q_n}{Q_{n+1}}\left(\frac{P_{n+2}}{Q_{n+2}}-\frac{P_{n+1}}{Q_{n+1}}\right)+\frac{Q_n}{Q_{n+1}}\left(\frac{P_{n+1}}{Q_{n+1}}-\frac{P_n}{Q_n}\right),$$ which takes the form $$U_{n+2}=\frac{-R_{n+1}}{x-2+R_{n+1}}U_{n+1}, \qquad \text{where} \qquad U_{n+1}=\frac{P_{n+1}}{Q_{n+1}}-\frac{P_n}{Q_n} \qquad \text{and} \qquad R_{n+1}=\frac{Q_n}{Q_{n+1}}.$$ I might write up the rest later, but for now here is roughly a useful item for checking $U_n$ is (or not) a Cauchy sequence: $$U_n=\pm\prod_{j=0}^{n-1}\frac{R_j}{x-2+R_j}.$$ The hope is to explore the story about the desired convergence of $r_n(x)$ $\dots$

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 and denominator polynomials of $r_n$, respectively. Then both polynomials share a common recurrence; namely, $$P_{n+2}+(x-2)P_{n+1}+P_n=0 \qquad \text{and} \qquad Q_{n+2}+(x-2)Q_{n+1}+Q_n=0.\tag1$$ They differ only in the initial condition where $P_0=1, P_1=1-x$ while $Q_0=1, Q_1=2-x$. The importance of such a description is that (1) the original recursive relations are decoupled here; (2) it is more amenable to an asymptotic analysis; (3) it reveals the roots being in $[0,4]$ due to the interlacing property of three-term recurrences.

Note. The original numerator and denominator differ by $\pm$ sign from $P_n$ and $Q_n$, but this makes no difference for the ratio $r_n$.

In fact (Fedor!), $$r_n=\frac{P_n(x)}{Q_n(x)}=\sqrt{x}\,\frac{U_{2n}(\sqrt{x}/2)}{U_{2n+1}(\sqrt{x}/2)}$$ where $U_n(y)$ are Chebyshev polynomials of the 2nd kind, expressible as $$U_n(y)=\frac{(y+\sqrt{y^2-1})^n-(y-\sqrt{y^2-1})^n}{2\sqrt{y^2-1}}.$$ If $y\geq1$, or equivalently $z=y+\sqrt{y^2-1}\geq1$, then $$\lim_{n\rightarrow\infty}\frac{U_n(y)}{U_{n+1}(y)}= \lim_{n\rightarrow\infty}\frac{z^n-z^{-n}}{z^{n+1}-z^{-n-1}}=\frac1z=y-\sqrt{y^2-1}.$$ If $0<y<1$ then the complex modulus $\vert z\vert=1$ and hence $\lim_{n\rightarrow\infty}\frac{U_n(y)}{U_{n+1}(y)}$ fails to exist.

If $y=0$ then apparently the limit is $0$.

When $y=\frac{\sqrt{x}}2$, the conditions become $$\lim_{n\rightarrow\infty}r_n(x)=\frac{x-\sqrt{x^2-4x}}2$$ if $x\geq4$ or $x=0$. Otherwise (if $0<x<4$) this limit does not exist for being oscillatory!

Finally, since the roots of Chebyshev polynomials $U_n(y)$ lie in $[-1,1]$ it follows that the roots of $U_n(\sqrt{x}/2)$ must be limited in the range $[0,4]$.

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 and denominator polynomials of $r_n$, respectively. Then both polynomials share a common recurrence; namely, $$P_{n+2}+(x-2)P_{n+1}+P_n=0 \qquad Q_{n+2}+(x-2)Q_{n+1}+Q_n=0.$$ They differ only in the initial condition where $P_0=1, P_1=1-x$ while $Q_0=1, Q_1=2-x$. The importance of such a description is that (1) the original recursive relations are decoupled here; (2) it is more amenable to an asymptotic analysis; (3) it reveals the roots being in $[0,4]$ due to the interlacing property of three-term recurrences.

Note. The original numerator and denominator differ by $\pm$ sign from $P_n$ and $Q_n$, but this makes no difference for the ratio $r_n$.

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 and denominator polynomials of $r_n$, respectively. Then both polynomials share a common recurrence; namely, $$P_{n+2}+(x-2)P_{n+1}+P_n=0 \qquad \text{and} \qquad Q_{n+2}+(x-2)Q_{n+1}+Q_n=0.\tag1$$ They differ only in the initial condition where $P_0=1, P_1=1-x$ while $Q_0=1, Q_1=2-x$. The importance of such a description is that (1) the original recursive relations are decoupled here; (2) it is more amenable to an asymptotic analysis; (3) it reveals the roots being in $[0,4]$ due to the interlacing property of three-term recurrences.

Note. The original numerator and denominator differ by $\pm$ sign from $P_n$ and $Q_n$, but this makes no difference for the ratio $r_n$.

In (1), multiply the 1st equation through by $Q_{n+2}$, the 2nd by $P_{n+2}$ and subtract to get $$(x-2)[P_{n+1}Q_{n+2}-Q_{n+1}P_{n+2}]=Q_nP_{n+2}-P_nQ_{n+2}.$$ Divide by $Q_{n+1}Q_{n+2}$ and rearrange a bit to get $$(x-2)\left(\frac{P_{n+1}}{Q_{n+1}}-\frac{P_{n+2}}{Q_{n+2}}\right) =\frac{Q_n}{Q_{n+1}}\left(\frac{P_{n+2}}{Q_{n+2}}-\frac{P_{n+1}}{Q_{n+1}}\right)+\frac{Q_n}{Q_{n+1}}\left(\frac{P_{n+1}}{Q_{n+1}}-\frac{P_n}{Q_n}\right),$$ which takes the form $$U_{n+2}=\frac{-R_{n+1}}{x-2+R_{n+1}}U_{n+1}, \qquad \text{where} \qquad U_{n+1}=\frac{P_{n+1}}{Q_{n+1}}-\frac{P_n}{Q_n} \qquad \text{and} \qquad R_{n+1}=\frac{Q_n}{Q_{n+1}}.$$ I might write up the rest later, but for now here is roughly a useful item for checking $U_n$ is (or not) a Cauchy sequence: $$U_n=\pm\prod_{j=0}^{n-1}\frac{R_j}{x-2+R_j}.$$ The hope is to explore the story about the desired convergence of $r_n(x)$ $\dots$

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 and denominator polynomials of $r_n$, respectively. Then both polynomials share a common recurrence; namely, $$P_{n+2}+(x-2)P_{n+1}+P_n=0 \qquad Q_{n+2}+(x-2)Q_{n+1}+Q_n=0.$$ They differ only in the initial condition where $P_0=1, P_1=1-x$ while $Q_0=1, Q_1=2-x$. The importance of such a description is that (1) the original recursive relations are decoupled here; (2) it is more amenable to an asymptotic analysis; (3) it reveals the roots being in $[0,4]$ due to the interlacing property of three-terms recurrences.

Note. The original numerator and denominator differ by $\pm$ sign from $P_n$ and $Q_n$, but this makes no difference for the ratio $r_n$.

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 and denominator polynomials of $r_n$, respectively. Then both polynomials share a common recurrence; namely, $$P_{n+2}+(x-2)P_{n+1}+P_n=0 \qquad Q_{n+2}+(x-2)Q_{n+1}+Q_n=0.$$ They differ only in the initial condition where $P_0=1, P_1=1-x$ while $Q_0=1, Q_1=2-x$. The importance of such a description is that (1) the original recursive relations are decoupled here; (2) it is more amenable to an asymptotic analysis; (3) it reveals the roots being in $[0,4]$ due to the interlacing property of three-term recurrences.

Note. The original numerator and denominator differ by $\pm$ sign from $P_n$ and $Q_n$, but this makes no difference for the ratio $r_n$.