TITLE: Proof $\sum_{k=0}^\infty \binom{k}{n-k} = f_{n+1}$ where $f_n$ is the n-th Fibonacci-number
QUESTION [5 upvotes]: In our combinatorics script it is written, that
$$\sum_{k=0}^\infty \binom{k}{n-k} = f_{n+1}$$ where $f_n$ is the n-th Fibonacci-number.
Apparently this can be proven through the generating function 
$$A(x) = \sum_{n=0}^\infty x^n \sum_{k=0}^\infty \binom{k}{n-k}$$
I've looked on stackexchange math and on the internet, but I couldn't find a proof.
I know that $$\sum_{k\ge0} \binom{n-k}k=f_{n+1}$$
and it can be proven through this
$$\sum_{k\ge0} \binom{n+1-k}k=\sum_{k\ge0} \binom{n-k}k + \sum_{k\ge0} \binom{n-k}{k-1}=
\sum_{k\ge0} \binom{n-k}k + \sum_{k\ge0} \binom{n-1-(k-1)}{k-1}=
\sum_{k\ge0} \binom{n-k}k + \sum_{j\ge0} \binom{n-1-j}{j}= f_{n+1}+f_n = f_{n+2}.$$
But I don't know how its done for the first case.

REPLY [7 votes]: \begin{align}
A(x)
&= \sum_{n\ge 0} x^n\sum_{k\ge 0} \binom{k}{n-k}
\\&= \sum_{k\ge 0} \sum_{n\ge 0} \binom{k}{n-k}x^n
\\&= \sum_{k\ge 0} x^k\sum_{n\ge k} \binom{k}{n-k}x^{n-k}
\\&= \sum_{k\ge 0} x^k\sum_{n\ge 0} \binom{k}{n}x^{n}
\\&= \sum_{k\ge 0} x^k(1+x)^k
\\&= \frac1{1-(x+x^2)}
\end{align}
All that remains is to recall $\sum_{n\ge 0} f_nx^n=\frac{x}{1-x-x^2}$.