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.

 A: $$
\begin{align}
\frac{\sum_{r=0}^n\binom{2n}{2r}2^r}{\sum_{r=0}^{n-1}\binom{2n}{2r}2^r}
&=\sqrt2\frac{\sum_{r=0}^n\binom{2n}{2r}\sqrt2^{2r}}{\sum_{r=0}^{n-1}\binom{2n}{2r+1}\sqrt2^{2r+1}}\\
&=\sqrt2\frac{\sum_{r=0}^{2n}\binom{2n}{r}\left(-\sqrt2\right)^r}{\sum_{r=0}^{n-1}\binom{2n}{2r+1}\sqrt2^{2r+1}}
+\sqrt2\\
&=\sqrt2\frac{\left(1-\sqrt2\right)^{2n}}{\sum_{r=0}^{n-1}\binom{2n}{2r+1}\sqrt2^{2r+1}}+\sqrt2\\[6pt]
&\to\sqrt2
\end{align}
$$

Another way to look at it:
$$
\sum_{r=0}^n\binom{2n}{2r}\sqrt2^{2r}-\sum_{r=0}^{n-1}\binom{2n}{2r+1}\sqrt2^{2r+1}=\left(1-\sqrt2\right)^{2n}
$$
and
$$
\sum_{r=0}^n\binom{2n}{2r}\sqrt2^{2r}+\sum_{r=0}^{n-1}\binom{2n}{2r+1}\sqrt2^{2r+1}=\left(1+\sqrt2\right)^{2n}
$$
so the ratio is
$$
\sqrt2\frac{\left(1+\sqrt2\right)^{2n}+\left(1-\sqrt2\right)^{2n}}{\left(1+\sqrt2\right)^{2n}-\left(1-\sqrt2\right)^{2n}}
=\sqrt2\frac{1+\left(\frac{-1}{3+2\sqrt2}\right)^{2n}}{1-\left(\frac{-1}{3+2\sqrt2}\right)^{2n}}
$$
which tends to $\sqrt2$ since $\left|\,\frac{1-\sqrt2}{1+\sqrt2}\,\right|=\frac1{3+2\sqrt2}\lt\frac15$.
A: Notice that for a polynomial $p(z)$  the expression
$$\left.\frac{1}{2} p(z) + \frac{1}{2} p(-z)\right|_{z=q}$$ 
gives the sum of the coefficients of even powers of $z$  times $q^k$ where $k$ is the summation index and
$$\left.\frac{1}{2} p(z) - \frac{1}{2} p(-z)\right|_{z=q}$$ 
the sum of the coefficients of odd powers of $z$ times $q^k$.
In the present case we have $p(z) = (1+z)^{2n}$ with $q=\sqrt{2}.$ Hence the limit is
$$\lim_{n\to\infty} q \frac{(1+q)^{2n}+(1-q)^{2n}}{(1+q)^{2n}-(1-q)^{2n}}
= \lim_{n\to\infty}q \left(1+ 2\frac{(1-q)^{2n}}{(1+q)^{2n}-(1-q)^{2n}}\right) \\ =
\lim_{n\to\infty}q \left(1+ 2\frac{1}{\left(\frac{1+q}{1-q}\right)^{2n}-1}\right) =q$$
because $$\left|\frac{1+q}{1-q}\right| > 1$$ when $q=\sqrt{2}.$
A: 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 $x=\sqrt2>0$
