For all diagonals $a$ of Pascal's triangle (figurate numbers), $\sum_{k=a}^\infty {k\choose a}\frac{1}{2^k}=2$? I am looking for the derivation of the closed form along any given diagonal $a$ of Pascal's triangle,
$$\sum_{k=a}^n {k\choose a}\frac{1}{2^k}=?$$
Numbered observations follow. As for the limit proposed in the title given by:
Observation 1
$$\sum_{k=a}^\infty {k\choose a}\frac{1}{2^k}=2,$$
when I calculate the sums numerically using MS Excel for any $a$ within the domain ($0\le a \le100$) the sum approaches 2.000000 in all cases within total steps $n\le285$. The first series with $a=0$ is a familiar geometric series, and perhaps others look familiar to you as well:
$$\sum_{k=0}^\infty {k\choose 0}\frac{1}{2^k}=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+... =2,$$
$$\sum_{k=1}^\infty {k\choose 1}\frac{1}{2^k}=\frac{1}{2}+\frac{1}{2}+\frac{3}{8}+\frac{1}{4}+\frac{5}{32}... =2,$$
$$\sum_{k=2}^\infty {k\choose 2}\frac{1}{2^k}=\frac{1}{4}+\frac{3}{8}+\frac{3}{8}+\frac{5}{16}+\frac{15}{64}... =2,$$
but it is both surprising and elegantly beautiful that these sums across all diagonals appear to approach the same value. Some additional observations from the numerically determined sums:
Observation 2
The maximum value of any term ${k\choose a}\frac{1}{2^k}$ within a diagonal $a$ for the domain $(a>0)$ is attained at $k=2a-1$ and repeated for the term immediately following ($k=2a$).
Observation 3
$$\sum_{k=a}^{2a} {k\choose a}\frac{1}{2^k}=1$$
Observation 4
$$\sum_{k=a}^{n} {k\choose a}\frac{1}{2^k} + \sum_{k=n-a}^{n} {k\choose n-a}\frac{1}{2^k}=2$$
It's very likely that the general closed form has been derived before, but searching for the past several days has produced no results. It appears that setting up the appropriate generating function may play a role, but I am at a loss as to how to proceed. Looking forward to the responses.

The generating function proof of Observation 1.
Write $$f(x,z)=\sum_{n=0}^{\infty}(1+x)^nz^n=\frac{1}{1-z(x+1)}$$ Then the coefficient of $x^a$ is $$h_a(z)=\sum_{k=a}^\infty \binom ka z^k$$
When $z=1/2,$ then $h_a(1/2)$ is your value.
But $$f(x,1/2)=\frac1{1/2-x/2}=\frac{2}{1-x}$$
So $h_a(1/2)=2,$ the result you want.
For any particular value $|z_0|<1,$ you get:
$$
\begin{align} 
f(x,z_0)&=\frac{1}{1-z_0}\frac1{1-x\frac{z_0}{1-z_0}}\\&=\sum_{n=0}^{\infty}\frac{z_0^n}{(1-z_0)^{n+1}}x^n
\end{align} 
$$
So $$h_a(z_0)=\frac{z_0^a}{(1-z_0)^{a+1}}\tag1$$

The probability proof of Observation 1.
Let $p_n$ be the probability that in an infinite sequence of tosses of a fair coin, that after $n$ toss you have $a$ heads and the next toss is a another heads. So $$p_n=\binom na \frac1{2^{n+1}}.$$ Since it is probability zero that you will never get $a+1$ heads, this means:
$$\sum_{n} p_n=1.$$
If the probality of heads is $p,$ then:
$$p_n=\binom na p^{a+1}(1-p)^{n-a}$$
This lets you get, with a little manipulation, the same result as $(1)$ with $z_0=1-p.$
Observation 2.
If $x_n=\binom na\frac1{2^n},$ then when $n\geq a,$  we get a recurrence:
$$x_{n+1}=\frac{n+1}{2(n+1-a)}x_n.$$
Show that if $a\leq n<2a-1,$ then $2(n+1-a)<n+1,$ so $x_{n+1}>x_n.$
When $n=2a-1,$ you get $2(n+1-a)=n+1.$ So $x_{2a-1}=x_{2a}.$
When $n>2a-1$ you likewise get $x_{n+1}<x_n.$
Observation 3.
$$\sum_{k=a}^{2a} \binom ka \frac1{2^{k+1}}\tag2$$ is the probability after $2a+1$ tosses of a fair coin, that you’ve gotten $a+1$ or more heads, with $k$ being the point where the $k+1$st toss was when you got the $a+1$st heads.
But $(2)$ is also equal to the probability that you got $a+1$ or more tails.
And you always get $a+1$ or more heads or $a+1$ or more tails in $2a+1$ tosses, but not both. So (2) is equal to $\frac12.$ Multiply by $2$ and you get your identity.
Observation 4.
This is roughly the same as for observation $(3).$
Let $b=n-a.$
Then $$\sum_{k=a}^{a+b} \binom ka \frac1{2^{k+1}}\tag3$$
is the probability in $a+b+1$ tosses of a fair coin that you get $a+1$ or more heads.
And:
$$\sum_{k=b}^{a+b}\binom kb\frac1{2^{k+1}}\tag4$$
is the probability of $b+1$ or more tails in $a+b+1$ tosses of a fair coin.
As with Observation 3, exactly one of the events counted in $(3)$ and $(4)$ is true, so the sum of these two values is $1.$
Multiplying that sum by $2$ gives you the Observation 4.

If $p$ is the probability of heads in a coin, you get a generalization of Observation 4:
$$p^{a+1}\sum_{i=0}^{b}\binom {a+i}a (1-p)^{k-a}+(1-p)^{b+1}\sum_{j=0}^{a}\binom {b+j}b p^{j}=1.$$
Another way to write the probability that you get more than $a$ heads win $a+b+1$ tosses is:
$$\sum_{k=a+1}^{a+b+1}\binom{a+b+1}{k}p^k(1-p)^{a+b+1-k}\\=p^{a+1}\sum_{j=0}^{b}\binom {a+b+1}jp^{b-j}(1-p)^j$$ where $k$ goes over all the possible numbers of heads $>a,$ and $j=a+b+1-k$ loops over the possible numbers of tails.
That gives a new identity, swapping $p$ and $1-p,$ you get:
$$\sum_{j=0}^{b}\binom {a+b+1}jp^j(1-p)^{b-j}= \sum_{i=0}^{b}\binom {a+i}a p^{i}$$
When $p=\frac12,$ this gives you: $$\frac{1}{2^{a+b}}\sum_{j=0}^b \binom{a+b+1}j=\sum_{k=a}^{a+b}\binom ka\frac{1}{2^k}.\tag 5$$
Indeed, the identity $(5)$ can be seen as the heart of the all the observations other than $2.$
For example, looking at the left side, we see it is $\frac{1}{2^{a+b}}\left(2^{a+b+1}-g(b)\right),$ where $g(b)=\sum_{i=0}^a\binom{a+b+1}i$ is a polynomial of degree $a.$ So we get as $b\to \infty,$ that the right side converges to $2.$
You can prove $(5)$ directly by induction.
When $b=0,$ it is easy.
$$
\begin{align} 
\sum_{j=0}^{b+1}\binom{a+b+2}j&=\sum_{j=0}^{b+1}\left(\binom{a+b+1}j+\binom{a+b+1}{j-1}\right)\\&=\binom{a+b+1}{b+1}+2\sum_{j=0}^b\binom {a+b+1}j\\&= \binom{a+b+1}{a}+2\sum_{j=0}^b\binom {a+b+1}j
\end{align}
$$ Dividing by $2^{a+b+1}$ on both sides, you get the left side for $(5)$ increases from $b$ to $b+1$ by $\frac1{2^{a+b+1}}\binom{a+b+1}{a}.$
I’m not sure the induction proof of $(5)$ is the most edifying - it is clearer if you understand the underlying probabilities. But it might be the most direct.
