Combinatoric formula summing one While working in a problem derived from a mathematical model of a game, I found the following formula:
$$
lim_{n\rightarrow\infty}\sum_{i=1}^m(-1)^{i-1} \binom{n-i}{i} c^{-2i}=1,
$$
where $c$ is any positive integer major or equal than 2, $m=n/2$ if $n$ is even and $m=\frac{n-1}{2}$ if $n$ is odd.
I know it is correct because I have tested it in the computer, but I have no idea of how to prove it.
 A: At this answer the following formula has been posted:
$$a^n+b^n=(a+b)^n-\sum _{i=1}^{n-1}(-1)^{i-1}\frac{n}{n-i}\binom{n-i}{i}(a+b)^{n-2i}(ab)^i$$
Factoring out $(a+b)^{n}$
$$a^n+b^n=(a+b)^n-(a+b)^{n}\cdot \sum _{i=1}^{n-1}(-1)^{i-1}\frac{n}{n-i}\binom{n-i}{i}(a+b)^{-2i}(ab)^i$$
$$(a+b)^n-a^n+b^n=(a+b)^{n}\cdot \sum _{i=1}^{n-1}(-1)^{i-1}\frac{n}{n-i}\binom{n-i}{i}(a+b)^{-2i}(ab)^i$$
Dividing the equation by $(a+b)^{n}$
$$1-\frac{a^n+b^n}{(a+b)^{n}}= \sum _{i=1}^{n-1}(-1)^{i-1}\frac{n}{n-i}\binom{n-i}{i}(a+b)^{-2i}(ab)^i$$
We can set $a\cdot b=1$. Consequently $(ab)^i=1$ disappears.
$$1-\frac{a^n+b^n}{(a+b)^{n}}= \sum _{i=1}^{n-1}(-1)^{i-1}\frac{n}{n-i}\binom{n-i}{i}(a+b)^{-2i}$$
Then we have two equations:

*

*$a\cdot b=1\Rightarrow a=\frac{1}b$

*$a+b=c$
The resulting quadratic equation $\frac1b+b=c$ has a solution for $b$ only if the discriminant is non-negative. This is when $c^2\geq 4$. If confirms the condition of Claude Leibovici. Now we can look at the limit.
$$1-\lim_{n \to \infty}\frac{a^n+b^n}{(a+b)^{n}}= \lim_{n \to \infty}
\sum _{i=1}^{n-1}(-1)^{i-1}\frac{n}{n-i}\binom{n-i}{i}c^{-2i}$$
$\lim\limits_{n \to \infty}\frac{n}{n-i}=1$
$$1-\lim_{n \to \infty}\frac{a^n+b^n}{(a+b)^{n}}= \lim_{n \to \infty}
\sum _{i=1}^{n-1}(-1)^{i-1}\binom{n-i}{i}c^{-2i}$$
$$1= \lim_{n \to \infty}
\sum _{i=1}^{n-1}(-1)^{i-1}\binom{n-i}{i}c^{-2i}$$
If we calculate at the limit ($\to$ infinity) we don´t have to distinguish between the odd- and the even-cases.
A: You are facing the problem of the limits of the gaussian hypergeometric functions.
If $n=2p$, the result of the summation is
$$1-\, _2F_1\left(\frac{1}{2}-p,-p;-2 p;\frac{4}{c^2}\right)$$
If $n=2p+1$, the result of the summation is
$$1-\, _2F_1\left(-p-\frac{1}{2},-p;-2 p-1;\frac{4}{c^2}\right)$$
But you need $c^2>4$
