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?

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 $x=\frac 12$
$$\sum_{k=n}^{2n}  \binom{k}{n} x^k=1+ \left(4 n+2-\frac{4\, \Gamma
   \left(n+\frac{3}{2}\right)}{\sqrt{\pi }\, \Gamma
   (n+1)}\right)\left(x-\frac{1}{2}\right)+O\left(\left(x-\frac{1}{2}\right)^2\right)$$
