Non-probabilistic/non-combinatoric proof of $\sum_{k=n}^{2n} \frac{{k \choose n}}{2^k}=1$ The summation   $$S_n=\sum_{k=n}^{2n} \frac{{k \choose n}}{2^k}=1~~~~(1)$$
has been proved using probabilistic/combinatoric method in MSE earlier:
Combinatorial or probabilistic proof of a binomial identity
Here we give an analytic proof for (1):
$$S_n=\sum_{k=n}^{2n} \frac{{k-1 \choose n-1}+{k-1 \choose n}}{2^k}~~~~~(2)$$
Let $k-1=p$, then
$$S_n=\sum_{p=n-1}^{2n-1} \frac{{p \choose n-1}+{p \choose n}}{2^{p+1}}~~~~(3)$$
Take the first term
$$\sum_{p=n-1}^{2n-1} \frac{{p \choose n-1}}{2^{p+1}}=\sum_{p=n-1}^{2n-2} \frac{{p \choose n-1}}{2^{p+1}}+\frac{{2n-1 \choose n-1}}{2^{2n-1}}=\frac{S_{n-1}}{2}+\frac{{2n-1 \choose n-1}}{2^{2n}}~~~~~(4)$$
Now take the second term
$$\sum_{p=n-1}^{2n-1} \frac{{p \choose n}}{2^{p+1}}=\sum_{p=n}^{2n} \frac{{p \choose n}}{2^{p+1}}+\frac{{n-1 \choose n}}{2^{n}}-\frac{{2n \choose n}}{2^{2n+1}}~~~~(5)$$
The second term in RHS vanishes and the third term is equal and opposite to the second term of the RHS of (4). Adding (4) and (5), we get
$$S_n=\frac{S_{n-1}}{2}+\frac{S_{n}}{2} \implies S_n=S_{n-1}$$So $S_n$ is a constant (independent of $n$) we can have $S_n=S_{n-1}=S_1=1$
The question is: What could be other analytic proofs of (1) without using the ideas of probability/combinatorics?
 A: We seek to show that
$$\sum_{k=n}^{2n} 2^{-k} {k\choose n} = 1$$
or alternatively
$$S_n = \sum_{k=0}^n 2^{-k} {k+n\choose n} = 2^n.$$
Using an Iverson bracket we find
$$\sum_{k\ge 0} 2^{-k} {k+n\choose n}
[z^n] \frac{z^k}{1-z}
= [z^n] \frac{1}{1-z}
\sum_{k\ge 0} 2^{-k} {k+n\choose n} z^k
\\ = [z^n] \frac{1}{1-z}
\frac{1}{(1-z/2)^{n+1}}.$$
This is
$$\mathrm{Res}_{z=0} \frac{1}{z^{n+1}} \frac{1}{1-z}
\frac{1}{(1-z/2)^{n+1}}
\\ = (-1)^n 2^{n+1} \mathrm{Res}_{z=0}
\frac{1}{z^{n+1}} \frac{1}{z-1}
\frac{1}{(z-2)^{n+1}}.$$
Residues sum to zero and the residue at infinity is zero by inspection.
Hence we take minus the sum of the residues at $z=1$ and $z=2.$ We
obtain for $z=1$
$$-(-1)^n 2^{n+1} (-1)^{n+1} = 2^{n+1}.$$
We write for $z=2$
$$- (-1)^n 2^{n+1} \mathrm{Res}_{z=2}
\frac{1}{((z-2)+2)^{n+1}} \frac{1}{(z-2)+1}
\frac{1}{(z-2)^{n+1}}
\\ = (-1)^{n+1} \mathrm{Res}_{z=2}
\frac{1}{(1+(z-2)/2)^{n+1}} \frac{1}{(z-2)+1}
\frac{1}{(z-2)^{n+1}}.$$
This yields
$$(-1)^{n+1} \sum_{q=0}^n (-1)^q 2^{-q} {n+q\choose n} (-1)^{n-q}
= - \sum_{q=0}^n 2^{-q} {n+q\choose n} = - S_n.$$
We have shown that $S_n = 2^{n+1} - S_n$ or $S_n = 2^n$ as claimed.
A: Using algebra.
$$\sum_{k=n}^{2n}  \binom{k}{n} x^k=(1-x)^{-(n+1)} x^n\Bigg[1-\binom{2 n+1}{n} ((1-x) x)^{n+1} \, _2F_1(1,2 n+2;n+2;x) \Bigg]$$
Developed as a series around $x=\frac 12$
$$\sum_{k=n}^{2n}  \binom{k}{n} x^k=1+ \left(4 n+2-\frac{4\, \Gamma
   \left(n+\frac{3}{2}\right)}{\sqrt{\pi }\, \Gamma
   (n+1)}\right)\left(x-\frac{1}{2}\right)+O\left(\left(x-\frac{1}{2}\right)^2\right)$$
