Find $\lim_{n\to\infty}$ of this quotient. Find, with proof, the value of this limit
$$\lim_{n\to\infty}\frac{\sum^n_{r=0}\binom{2n}{2r}\cdot2^r}{\sum^{n-1}_{r=0}\binom{2n}{2r+1}\cdot2^r}$$

I have tried using binomial identities but two problems occur:


*

*Only even binomial coefficients in numerator and only odd in denominator.

*The binomial coefficient occurs with $2^r$ and not with $2^{2r}$ which would be the binomial identity.


Generalization :
Using Binomial Theorem,
$$(1+x)^{2n}=1+\binom{2n}1x+\binom{2n}2x^2+\cdots+\binom{2n}{2n-1}x^{2n-1}+x^{2n}$$
$$\implies (1-x)^{2n}=1-\binom{2n}1x+\binom{2n}2x^2+\cdots-\binom{2n}{2n-1}x^{2n-1}+x^{2n}$$
$$\implies (1+x)^{2n}+(1-x)^{2n}=2\{1+\binom{2n}2x^2+\binom{2n}4x^4+\cdots+\binom{2n}{2n-2}x^{2n-2}+x^{2n}\}$$
$$\implies (1+x)^{2n}-(1-x)^{2n}=2\{\binom{2n}1x+\binom{2n}3x^3+\cdots+\binom{2n}{2n-3}x^{2n-3}+\binom{2n}{2n-1}x^{2n-1}\}$$
On division, $$\frac{\sum_{0\le r\le n}\binom{2n}{2r}x^{2r}}{x\sum_{0\le r\le n-1}\binom{2n}{2r+1}x^{2r}}=\frac{(1+x)^{2n}+(1-x)^{2n}}{(1+x)^{2n}-(1-x)^{2n}}=\frac{1+\left(\frac{1-x}{1+x}\right)^{2n}}{1-\left(\frac{1-x}{1+x}\right)^{2n}}$$ 
This will  $\to 1$ when $n\to\infty$ if $-1<\frac{1-x}{1+x}<1$
Now, $-1<\frac{1-x}{1+x}\iff 0<\frac{1-x}{1+x}+1=\frac2{1+x}\iff 1+x>0\iff x>-1\ \ \ \ (1)$
$\frac{1-x}{1+x}<1\iff \frac{1-x}{1+x}-1<0\iff \frac{-2x}{1+x}<0$
$\iff x(x+1)>0$ multiplying by $-\frac{(x+1)^2}2$ which is $<0$
$\implies $ either $(x>0$ and $x+1>0\iff x>-1)\implies x>0\ \ \ \ (2)$
or $(x<0$ and $x+1<0\iff x<-1)\implies x<-1$ which contradicts $(1)$
Combining $(1),(2)$ we need $x>0$
Here $x=\sqrt2>0$
