Let $a_n = \sum\limits_{i=\lceil \frac{n}{2}\rceil}^\infty\binom{2i}{n}\frac{1}{2^i}$. Prove that $a_{n+1} = 2a_n + a_{n-1}$.
I have tried considering the derivatives of $\frac{1}{1-x^2}=1+x^2+x^4+...$, and although this may work, it certainly does not seem like it would be the best way to go about solving this problem.
First, as Steven Stadnicki's question comment states, you can also get the same result as what you're asking for by starting "from $i = 0$" because "$\binom{a}{b} = 0$ when $a \lt b$" (note $\binom{a}{b}$ means the number of ways to choose $b$ items from among $a$ items, but there are $0$ ways to do this if $b \gt a$).
Next, for positive natural numbers $n$ and $k$, Pascal's rule states
$$\binom{n}{k} = \binom{n - 1}{k} + \binom{n - 1}{k - 1} \tag{1}\label{eq1A}$$
With positive natural numbers $i$ and $n$, this gives
$$\binom{2i}{n + 1} = \binom{2i - 1}{n + 1} + \binom{2i - 1}{n} \tag{2}\label{eq2A}$$
Using \eqref{eq1A} on both terms of the RHS of \eqref{eq2A} gives
$$\binom{2i - 1}{n + 1} = \binom{2i - 2}{n + 1} + \binom{2i - 2}{n} \tag{3}\label{eq3A}$$
$$\binom{2i - 1}{n} = \binom{2i - 2}{n} + \binom{2i - 2}{n - 1} \tag{4}\label{eq4A}$$
Substituting these into \eqref{eq2A}, multiplying both sides by $\frac{1}{2^{i}}$, summing from $i = 1$ to $\infty$, changing the summation indices on the RHS and making a few algebraic manipulations, you end up with
$$\begin{equation}\begin{aligned} \binom{2i}{n + 1} & = \binom{2i - 2}{n + 1} + 2\binom{2i - 2}{n} + \binom{2i - 2}{n - 1} \\ \binom{2i}{n + 1}\frac{1}{2^i} & = \frac{1}{2}\binom{2i - 2}{n + 1}\frac{1}{2^{i-1}} + \binom{2i - 2}{n}\frac{1}{2^{i-1}} + \frac{1}{2}\binom{2i - 2}{n - 1}\frac{1}{2^{i-1}} \\ \sum_{i=1}^{\infty}\binom{2i}{n + 1}\frac{1}{2^i} & = \frac{1}{2}\sum_{i=1}^{\infty}\binom{2(i - 1)}{n + 1}\frac{1}{2^{i-1}} + \sum_{i=1}^{\infty}\binom{2(i - 1)}{n}\frac{1}{2^{i-1}} + \frac{1}{2}\sum_{i=1}^{\infty}\binom{2(i - 1)}{n - 1}\frac{1}{2^{i-1}} \\ a_{n+1} & = \frac{1}{2}\sum_{i=0}^{\infty}\binom{2i}{n + 1}\frac{1}{2^{i}} + \sum_{i=0}^{\infty}\binom{2i}{n}\frac{1}{2^{i}} + \frac{1}{2}\sum_{i=0}^{\infty}\binom{2i}{n - 1}\frac{1}{2^{i}} \\ a_{n+1} & = \left(\frac{1}{2}\right)a_{n+1} + a_{n} + \left(\frac{1}{2}\right)a_{n-1} \\ \left(\frac{1}{2}\right)a_{n+1} & = a_{n} + \left(\frac{1}{2}\right)a_{n-1} \\ a_{n+1} & = 2a_n + a_{n-1} \end{aligned}\end{equation}$$
